The area of the circle in polar coordinates is 16π square units.
What is the area of a circle represented in polar coordinates?To calculate the area of the circle in polar coordinates, we can use the following steps:
Step 1: Convert the equation of the circle from rectangular coordinates to polar coordinates.
In polar coordinates, the conversion formulas are:
x = r * cos(θ)
y = r * sin(θ)
Given the equation of the circle as r = 8 * cos(θ), we can rewrite it in polar coordinates as:
r = 8 * cos(θ)
Step 2: Determine the limits of integration for θ.
The limits of integration for θ will depend on the range of values that allow the circle to be fully traced out.
Since r = 8 * cos(θ), the maximum value of r occurs when cos(θ) is at its minimum value, which is -1.
Therefore, the circle is fully traced out when θ ranges from 0 to π.
Step 3: Set up the integral to calculate the area.
The area of a circle in polar coordinates is given by the formula:
A = ∫[r(θ)]² * (1/2) dθ
Plugging in r = 8 * cos(θ), and the limits of integration for θ as 0 to π, we get:
A = ∫[8 * cos(θ)]² * (1/2) dθ from θ = 0 to θ = π
Simplifying, we get:
A = (1/2) * ∫[64 * cos²(θ)] dθ from θ = 0 to θ = π
Step 4: Evaluate the integral and calculate the area.
Usingen trigonometric idtity, cos²(θ) = (1 + cos(2θ))/2, we can rewrite the integral as:
A = (1/2) * ∫[64 * (1 + cos(2θ))/2] dθ from θ = 0 to θ = π
Simplifying further, we get:
A = (1/4) * ∫[64 + 64 * cos(2θ)] dθ from θ = 0 to θ = π
Now we can integrate term by term:
A = (1/4) * [64θ + 32 * sin(2θ)] from θ = 0 to θ = π
Plugging in the limits of integration, we get:
A = (1/4) * [64π + 32 * sin(2π)] - (1/4) * [0 + 32 * sin(0)]
Since,
sin(0) = 0 and sin(2π) = 0, we can simplify further:
A = (1/4) * 64π
Finally, we can simplify and express the area in terms of π:
A = 16π
So, the area of the circle with the equation r = 8 * cos(θ) in polar coordinates is 16π square units.
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For each of the following linear operators L on R3, find a matrix ,A such that L(x) = Ax for every x inR3. L((x1, x2, x3)T) = (x1, x1 + x2, x1 + x2 + x3)T) L((x1, x2 + 3x1, 2x1 - x3))T
The matrix representation of L is:
| 1 0 0 |
| 0 1 1 |
| 3 1 -1 |
To find the matrix representation of a linear operator L, we need to find the image of the standard basis vectors under L and then form a matrix from the resulting vectors.
For the first linear operator L, we have:
L((1,0,0)T) = (1,1,1)T
L((0,1,0)T) = (0,1,1)T
L((0,0,1)T) = (0,0,1)T
Therefore, the matrix representation of L is:
| 1 0 0 |
| 1 1 0 |
| 1 1 1 |
To check that this matrix represents L, we can multiply it by an arbitrary vector x = (x1, x2, x3)T:
[tex]| 1 0 0 | | x1 | | x1 |[/tex]
| 1 1 0 | x | x2 | = | x1+x2 |
| 1 1 1 | | x3 | | x1+x2+x3 |
which matches the formula for L(x) given in the problem.
For the second linear operator L, we have:
L((1,0,0)T) = (1,0,0)T
L((0,1,0)T) = (0,1,0)T + 3(1,0,0)T = (0,1,0)T + (3,0,0)T = (3,1,0)T
L((0,0,1)T) = 2(1,0,0)T - (0,0,1)T = (2,0,-1)T
Therefore, the matrix representation of L is:
| 1 0 0 |
| 0 1 1 |
| 3 1 -1 |
To check that this matrix represents L, we can multiply it by an arbitrary vector x = (x1, x2, x3)T:
[tex]| 1 0 0 | | x1 | | x1 |[/tex]
[tex]| 0 1 1 | x | x2 | = | x2 + x3 |[/tex]
| 3 1 -1 | | x3 | | 3x1 + x2 - x3 |
which matches the formula for L(x) given in the problem.
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Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value of x to be less than 0.001 rx) = sin(x), approximate f(0.7)
Answer:
herefore, the degree of the Maclaurin polynomial required for the error in the approximation of sin(x) at x = 0.7 to be less than 0.001 is 3. Using the Maclaurin series up to degree 3, we get:sin(0.7) ≈ 0.7 - 0.7^3/3!sin(0.7) ≈ 0.6433This approximation is accurate to within 0.001.
Step-by-step explanation:
We can use Taylor's theorem with the remainder in Lagrange form to estimate the error in approximating sin(x) with its Maclaurin polynomial:|Rn(x)| ≤ M * |x - a|^(n+1) / (n+1)!where:
Rn(x) is the remainder (the difference between the exact value of the function and its approximation using the Maclaurin polynomial)
M is an upper bound on the (n+1)st derivative of the function on the interval [0, x]
a is the center of the Maclaurin series (in this case, a = 0)
n is the degree of the Maclaurin polynomialSince sin(x) is continuous and differentiable for all x, we know that the Maclaurin series for sin(x) converges to sin(x) for all x. Therefore, we can use the Maclaurin series for sin(x) to approximate sin(0.7):sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...sin(0.7) ≈ 0.7 - 0.7^3/3! + 0.7^5/5!sin(0.7) ≈ 0.6442 (rounded to four decimal places)To find the degree of the Maclaurin polynomial required for the error in this approximation to be less than 0.001, we need to solve the following inequality for n:0.7^(n+1) / (n+1)! ≤ 0.001We can use a calculator or a table of values for factorials to solve this inequality. One possible method is to try different values of n until we find the smallest value that satisfies the inequality.Starting with n = 2, we get:0.7^3 / 3! ≈ 0.082This is not less than 0.001, so we try n = 3:0.7^4 / 4! ≈ 0.005This is less than 0.001, so we have found the degree of the Maclaurin polynomial required for the error to be less than 0.001:n = 3Therefore, the degree of the Maclaurin polynomial required for the error in the approximation of sin(x) at x = 0.7 to be less than 0.001 is 3. Using the Maclaurin series up to degree 3, we get:sin(0.7) ≈ 0.7 - 0.7^3/3!sin(0.7) ≈ 0.6433This approximation is accurate to within 0.001.
We need at least a degree 4 Maclaurin polynomial to approximate sin(x) at x = 0.7 with an error less than 0.001.
To determine the degree of the Maclaurin polynomial required for the error in the approximation of the function sin(x) at x = 0.7 to be less than 0.001, we need to consider the following:
1. The Maclaurin series for sin(x) is given by:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
2. The error in a Maclaurin series approximation can be estimated using the remainder term formula:
|error| ≤ |(x^n+1)/(n+1)!|
3. Plug in the desired error and x value (0.001 and 0.7, respectively) to find the smallest n such that the error is less than 0.001:
|0.001| ≤ |(0.7^n+1)/(n+1)!|
4. Iterate through different values of n (starting with n = 0) until the inequality is satisfied. Remember that n must be an even number as sin(x) is an odd function.
After iterating through different values of n, you will find that the smallest even n that satisfies the inequality is 4. Therefore, the degree of the Maclaurin polynomial required for the error in the approximation of sin(0.7) to be less than 0.001 is 4.
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find the inverse laplace transform of 8s 2s2−25s>5
The value of the function 8s/ 2s^2−25s using inverse Laplace transform is equal to 4e^(25t/2).
Function is equal to,
8s/ 2s^2−25s
Value of 's' after factorizing the denominator we get,
2s^2−25s = 0
⇒ s( 2s -25 ) =0
⇒ s =0 or s =25/2
Now apply partial fraction decomposition we get,
8s/ 2s^2−25s = A/s + B /(2s -25)
Simplify it we get,
⇒ 8s = A(2s -25) + Bs
Now substitute s =0 we get,
⇒ 0 = A (-25) + 0
⇒ A =0
and s = 25/2
⇒8(25/2) = A(2×25/2 -25 ) + B(25/2)
⇒100 = B(25/2)
⇒B = 8
Now ,
8s/ 2s^2−25s = 0/s + 8 /(2s -25)
⇒ 8s/ 2s^2−25s = 8 /(2s -25)
Take inverse Laplace transform both the side we get,
L⁻¹ [8s / (2s^2 - 25s)] = L⁻¹ [8/(2s - 25)]
Apply , L⁻¹ [1/(as + b)] = (1/a)e^(-bt/a),
here,
a = 2 , b = -25
L⁻¹ [8s / (2s^2 - 25s)]
= L⁻¹ [8/(2s - 25)]
= (8/2) e^(25t/2)
= 4e^(25t/2)
Therefore, the value of inverse Laplace transform for the given function is equal to 4e^(25t/2)
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The above question is incomplete, the complete question is:
Find the inverse Laplace transform of 8s/ 2s^2−25s.
In Chapter, we examined a picture of winning time in men’s 500meter speed skating plotted across time. The data represented in the plot started in 1924 and went through 2010. A regression equation relating winning time and year for 1924 to 2006 iswinning time = 273.06 - (0.11865)(year)a. Would the correlation between winning time and year be positive or negative? Explain.b. In 2010, the actual winning time for the gold medal was 34.91 seconds. Use the regression equation to predict the winning time for 2010, and compare the prediction to what actually happened. Was the actual winning time higher or lower than the predicted time?c. Explain what the slope of -0.11865 indicates in terms of how winning times change from year to year.
a. The correlation between winning time and year would be negative because the regression equation has a negative slope (-0.11865).The slope of -0.11865, actual winning time in 2010 was 34.91 seconds.
b. Using the regression equation, we can predict the winning time for 2010 as follows:
winning time = [tex]273.06 - (0.11865)(2010)[/tex]
winning time = [tex]273.06 - 239.2465[/tex]
winning time = [tex]33.8135 seconds[/tex]
The actual winning time in 2010 was 34.91 seconds, which is higher than the predicted time.
c. The slope of -0.11865 indicates that winning times decrease by an average of 0.11865 seconds per year. In other words, for each year that passes, the winning time decreases by approximately 0.12 seconds on average. This suggests that athletes are improving and getting faster over time, which is a common trend in many sports.
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After updating preferences, what is the monopolist's probability of obtaining a promising result, P(Promising) =Question 49 options:a) 17/40b) 18/40c) 19/40d) 20/40
After updating preferences, the monopolist's probability of obtaining a promising result is P(Promising) = 17/40. So, the answer is option a.17/40
To answer this question, we need to calculate the probability that the firm will obtain a promising result, given that it has updated its preferences.
If the firm chooses not to invest in R&D, its profit is given by:
π0 = (P - C(q))q = (12 - Q - 5 - 6Q)Q = Q(7 - 7Q)
If the firm invests in R&D, its profit is given by:
π1 = P(Promising) [P(Successful) * (12 - Q - 5 - 2Q) + P(Failure) * (12 - Q - 5 - 6Q)] - 4 + [1 - P(Promising)](12 - Q - 5 - 6Q)
where P(Promising) is the probability of obtaining a promising result, P(Successful) is the probability of the new technology being successful, and P(Failure) is the probability of the new technology being a failure.
Simplifying the equation above, we get:
π1 = P(Promising) [3/8 * (7 - Q) + 5/8 * (7 - 5Q)] - Q - 1
To determine whether the firm should invest in R&D or not, we need to compare the profits under the two scenarios.
If π0 > π1, the firm should not invest in R&D. If π0 < π1, the firm should invest in R&D. If π0 = π1, the firm is indifferent between the two options.
Setting π0 = π1, we can solve for Q and obtain the threshold quantity, Q*.
Q* = 9/4
If Q* > 0, the firm will invest in R&D. Otherwise, it will not invest in R&D.
Substituting Q* into the two profit functions, we obtain:
π0 = Q*(7 - 7Q*) = 81/16
π1 = P(Promising) [3/8 * (7 - Q*) + 5/8 * (7 - 5Q*)] - Q* - 1
Substituting the values given in the question, we obtain:
π1 = P(Promising) [21/8 - 3/8Q* - 15/8] - Q* - 1
Simplifying the equation above, we get:
π1 = P(Promising) [-3/8Q* + 6/8] - Q* + 1/8
π1 = P(Promising) [-27/32] - 17/32
Setting π0 = π1, we can solve for P(Promising) and obtain:
P(Promising) = 17/40
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exercise 2.1.2. show that y=ex and y=e2x are linearly independent.
the Wronskian W(y1, y2) is not identically zero, we can conclude that the functions y1(x) = e^x and y2(x) = e^(2x) are linearly independent.
To show that y=e^x and y=e^(2x) are linearly independent, we'll use the Wronskian test. The Wronskian is a determinant that helps determine the linear independence of two functions. For our functions y1(x) = [tex]e^x[/tex] and y2(x) = [tex]e^{2x),[/tex]the Wronskian is given by:
W(y1, y2) = [tex]\left[\begin{array}{ccc}y_1&y_2\\y'_1&y'_2\\\end{array}\right][/tex]
Now, we'll compute the derivatives and populate the matrix:
[tex]y_1'(x) = e^x\\y_2'(x) = 2e^{2x}[/tex]
W(y1, y2) =[tex]e^x2e^{2x}-e^xe^{2x}[/tex]
Next, we'll compute the determinant of this matrix:
[tex]W(y1, y2) = (e^x * 2e^{2x)}) - (e^x * e^{2x}))\\W(y1, y2) = e^{3x)} (2 - 1)\\W(y1, y2) = e^{3x}\\\\[/tex]
Since the Wronskian W(y1, y2) is not identically zero, we can conclude that the functions [tex]y1(x) = e^x[/tex]and [tex]y2(x) = e^{2x}[/tex] are linearly independent.
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we will now write a function that is the product of our two numbers, where x represents the smaller number and x + 56 represents the larger number as follows. f(x) = x(x + __ )
= x^2 + ( __ ) x
according to an article, there were 1,008,329 associate degrees awarded by u.s. community colleges in a certain academic year. a total of 612,034 of these degrees were awarded to women.. (Round your answers to three decimal places.) (a) If a person who received a degree in this year was selected at random, what is the probability that the selected student will be female? (b) What is the probability that the selected student will be male?
The answers of a, and b are the probability that the selected student will be female is approximately 0.607, and the probability that the selected student will be male is approximately 0.393.
(a) To find the probability that a randomly selected student will be female, we can use the following formula: P (female) = (number of degrees awarded to women) / (total number of associate degrees awarded).
P(female) = 612,034 / 1,008,329
P(female) ≈ 0.607 (rounded to three decimal places)
So, the probability that the selected student will be female is approximately 0.607.
(b) To find the probability that a randomly selected student will be male, we first need to determine the number of degrees awarded to men: (total number of associate degrees awarded) - (number of degrees awarded to women).
Degrees awarded to men = 1,008,329 - 612,034 = 396,295
Now, we can use the same formula as before: P(male) = (number of degrees awarded to men) / (total number of associate degrees awarded).
P(male) = 396,295 / 1,008,329
P(male) ≈ 0.393 (rounded to three decimal places)
So, the probability that the selected student will be male is approximately 0.393.
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The answers of a, and b are the probability that the selected student will be female is approximately 0.607, and the probability that the selected student will be male is approximately 0.393.
(a) To find the probability that a randomly selected student will be female, we can use the following formula: P (female) = (number of degrees awarded to women) / (total number of associate degrees awarded).
P(female) = 612,034 / 1,008,329
P(female) ≈ 0.607 (rounded to three decimal places)
So, the probability that the selected student will be female is approximately 0.607.
(b) To find the probability that a randomly selected student will be male, we first need to determine the number of degrees awarded to men: (total number of associate degrees awarded) - (number of degrees awarded to women).
Degrees awarded to men = 1,008,329 - 612,034 = 396,295
Now, we can use the same formula as before: P(male) = (number of degrees awarded to men) / (total number of associate degrees awarded).
P(male) = 396,295 / 1,008,329
P(male) ≈ 0.393 (rounded to three decimal places)
So, the probability that the selected student will be male is approximately 0.393.
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Express y +11+7-9 - 3y in the simplest form.
Answer:
-2y + 9
Step-by-step explanation:
Combine like terms: y - 3y = -2y
Combine the constant terms: 11 + 7 - 9 = 9
Put the combined terms together: -2y + 9
Please help if you can
Answer:
y = 1
Step-by-step explanation:
the equation of a horizontal line is
y = c ( c is the value of the y- coordinates the line passes through )
the line passes through (- 5, 1 ) with y- coordinate 1 , then
y = 1 ← equation of horizontal line
Answer: C) y=1
Step-by-step explanation: It couldn't be B or D, seeing as they are both vertical lines. We are left with y=-5 and y=1. The only points that y=-5 pass through have a y-coordinate of -5, and the point in question has a y-coordinate of 1. Your answer is C! Hope this helped.
How many 5-element subsets of s = {1, 2, 3, 4, 5, 6, 7, 8, 9} have more odd numbers than even numbers?
The number of 5-element subsets of s = {1, 2, 3, 4, 5, 6, 7, 8, 9} with more odd numbers than even numbers is 126.
To determine this, we will find the subsets that have either 3 or 5 odd numbers. First, consider the subsets with 3 odd numbers and 2 even numbers.
There are 5 odd numbers (1, 3, 5, 7, 9) and 4 even numbers (2, 4, 6, 8) in the set. So, we need to choose 3 odd numbers out of 5 and 2 even numbers out of 4. Using the combination formula, we get C(5, 3) * C(4, 2) = 10 * 6 = 60.
Next, consider the subsets with 5 odd numbers and 0 even numbers. In this case, we need to choose all 5 odd numbers and no even numbers. Using the combination formula, we get C(5, 5) * C(4, 0) = 1 * 1 = 1.
Finally, add the results from the two cases to get the total number of subsets: 60 + 1 = 126.
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Find the exact values of the sine, cosine, and tangent of the angle.
11π/12 = 3π/4 + π/6.
how will I determine
Sin 11π/12, Cos 11π/12, and Tan 11π/12?
The exact values of sine, cosine, and tangent of 11π/12 are:
sin(11π/12) = -√6/4 - √2/2
cos(11π/12) = -√6/4 + √2/2
tan(11π/12) = (√2 - √6) / 2
To determine the exact values of sine, cosine, and tangent of 11π/12, we first use the sum formula for sine and cosine:
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
cos(A + B) = cos(A) cos(B) - sin(A) sin(B)
In this case, we have:
11π/12 = 3π/4 + π/6
So, we can rewrite this as:
sin(11π/12) = sin(3π/4 + π/6)
cos(11π/12) = cos(3π/4 + π/6)
Using the sum formula, we get:
sin(11π/12) = sin(3π/4) cos(π/6) + cos(3π/4) sin(π/6) = (-√2/2)(√3/2) + (-√2/2)(1/2) = -√6/4 - √2/2
cos(11π/12) = cos(3π/4) cos(π/6) - sin(3π/4) sin(π/6) = (-√2/2)(√3/2) - (-√2/2)(1/2) = -√6/4 + √2/2
tan(11π/12) = sin(11π/12) / cos(11π/12) = (-√6/4 - √2/2) / (-√6/4 + √2/2) = (√2 - √6) / 2
Therefore, the exact values of sine, cosine, and tangent of 11π/12 are:
sin(11π/12) = -√6/4 - √2/2
cos(11π/12) = -√6/4 + √2/2
tan(11π/12) = (√2 - √6) / 2
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For this problem, use equation 4 from 9.1 and Toricelli's Law:
y=(y)(y)(y)=−2y‾‾‾‾√dydt=Bv(y)A(y)v(y)=−2gy
where g is about 9.8 ms29.8 ms2.
At =0t=0, a conical tank of height 225 cm225 cm and top radius 75 cm75 cm is filled with water. Water leaks through a hole in the bottom of area 2.2 cm22.2 cm2. Let y()y(t)be the water level at time t.
(a.) Show that the tank's cross-sectional area at height yy is (y)=19y2A(y)=19πy2. There is no answer to enter into WeBWorK for this part, but you must do this in order to move on.
(b.) Find a differential equation for y()y(t) and solve it.
y()y(t) =
(c.) How long does it take for the tank to empty? You can answer in seconds (s), minutes (min), or hours (hr)
t
It takes approximately 22.4 seconds for the tank to empty.
(a) To find the cross-sectional area of the tank at height y, we note that the tank is conical and use the formula for the area of a circle with radius r: A = πr^2. Since the radius of the tank varies with y, we express it in terms of y using similar triangles:
y / (225 cm) = r / (75 cm)
r = (y/225) * (75 cm)
Substituting this expression for r into the formula for the area, we get:
A(y) = π[(y/225) * (75 cm)]^2
= π(1/3) * y^2 / 4
Simplifying, we get:
A(y) = (π/12) * y^2 / 2
= (π/24) * y^2
Using this expression for A(y), we can write the differential equation for y(t).
(b) Taking the time derivative of the given equation and substituting in A(y) from part (a), we get:
d/dt [y^(3/2)] = -2g(π/24)y^2 / 2.2 cm^2
Simplifying and solving for dy/dt, we get:
dy/dt = - (4/3) * (g/2.2 cm^2) * y^(1/2)
This is a separable differential equation that can be solved by separating the variables and integrating:
∫ y^(-1/2) dy = - (4/3) * (g/2.2 cm^2) ∫ dt
2√y = (4/3) * (g/2.2 cm^2) * t + C
where C is the constant of integration. To determine C, we use the initial condition y(0) = 225 cm:
2√225 = (4/3) * (g/2.2 cm^2) * 0 + C
C = 30 cm
Substituting C into the equation above, we get:
2√y = (4/3) * (g/2.2 cm^2) * t + 30 cm
Squaring both sides and simplifying, we get:
y = [(3/4) * (2.2 cm^2/g)]^2 * (t - (4/3) * (2.2 cm^2/g) * 30 cm)^2
(c) The tank will empty when y = 0. Solving for t, we get:
t = (4/3) * (2.2 cm^2/g) * 30 cm
t ≈ 22.4 s
Therefore, it takes approximately 22.4 seconds for the tank to empty.
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how many ways can a person toss a coin 11 times so that the number of heads is between 7 and 9 inclusive?
A person can toss a coin 11 times in 470 or 471 ways so that the number of heads is between 7 and 9 inclusive
To solve this problem, we can use the binomial distribution formula to find the probability of getting 7, 8, or 9 heads in 11 tosses of a fair coin. Then we can sum up these probabilities to get the total number of ways to get between 7 and 9 heads.
The binomial distribution formula is:
[tex]P(X = k) = C(n, k)[/tex]× [tex]p^k[/tex]× [tex](1 - p)^{n - k}[/tex]
where:
P(X = k) is the probability of getting k heads in n tosses of a coin
C(n, k) is the number of combinations of n items taken k at a time, which is given by [tex]C(n, k) = n! / (k![/tex] × [tex](n - k)!)[/tex]
p is the probability of getting a head on one toss of the coin (since the coin is fair, p = 0.5)
(1 - p) is the probability of getting a tail on one toss of the coin
Using this formula, we can find the probabilities of getting 7, 8, or 9 heads in 11 tosses:
[tex]P(X = 7) = C(11, 7)[/tex] × [tex]0.5^7[/tex] × [tex]0.5^4 = 330[/tex] × [tex]0.0078[/tex] × [tex]0.0625 = 0.1613[/tex]
[tex]P(X = 8) = C(11, 8)[/tex] × [tex]0.5^8[/tex] × [tex]0.5^3 = 165[/tex]× [tex]0.0039[/tex] × [tex]0.125 = 0.0557[/tex]
[tex]P(X = 9) = C(11, 9)[/tex] × [tex]0.5^9[/tex] × [tex]0.5^2 = 55[/tex] × [tex]0.00195[/tex] × [tex]0.25 = 0.0127[/tex]
To get the total probability of getting between 7 and 9 heads, we can add up these probabilities:
[tex]P(7 < = X < = 9) = P(X = 7) + P(X = 8) + P(X = 9) = 0.1613 + 0.0557 + 0.0127 = 0.2297[/tex]
Therefore, the probability of getting between 7 and 9 heads in 11 tosses of a fair coin is 0.2297. To find the number of ways to get between 7 and 9 heads, we can multiply this probability by the total number of possible outcomes, which is[tex]2^11 = 2048[/tex]:
Number of ways[tex]= 0.2297[/tex] × [tex]2048 = 470.9[/tex]
Since we can't have a fraction of a way, the actual number of ways to get between 7 and 9 heads is either 470 or 471. Therefore, a person can toss a coin 11 times in 470 or 471 ways so that the number of heads is between 7 and 9 inclusive.
To count the number of ways to toss a coin 11 times so that the number of heads is between 7 and 9 inclusive, we need to count the number of outcomes that have exactly 7, 8, or 9 heads.
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a researcher reports t(30) = 6.35, p < .01 for an independent-measures experiment. calculate the effect size measure (r2).
Effect size measure (r²) for this independent-measures experiment is 0.412.
How to calculate the effect size measure (r²) for an independent-measures t-test?We first need to find the value of t and the degrees of freedom (df).
From the information given, t(30) = 6.35, which means that the t-value is 6.35 and the degrees of freedom are 30.
We can use the following formula to calculate r²:
r² = t² / (t² + df)
Plugging in the values we have:
r² = (6.35)² / [(6.35)² + 30] = 0.412
Therefore, the effect size measure (r²) for this independent-measures experiment is 0.412. This indicates a large effect size.
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What is the 97% confidence interval for a sample of 204 soda cans that have a mean amount of 12.05 ounces and a standard deviation of 0.08 ounces?(12.038, 12.062)(11.970, 12.130)(11.970, 12.130)(12.033, 12.067)
The option: (12.038, 12.062)
How to calculate the 97% confidence interval?Hi, I'd be happy to help you calculate the 97% confidence interval for the given data. To find the 97% confidence interval for a sample of 204 soda cans with a mean amount of 12.05 ounces and a standard deviation of 0.08 ounces, follow these steps:
1. Identify the sample size (n), mean (µ), and standard deviation (σ): n = 204, µ = 12.05, σ = 0.08
2. Determine the confidence level, which is 97%. To find the corresponding z-score, you can use a z-table or calculator. The z-score for 97% confidence is approximately 2.17.
3. Calculate the standard error (SE) using the formula: SE = σ / √n. In this case, SE = 0.08 / √204 ≈ 0.0056.
4. Multiply the z-score by the standard error to find the margin of error (ME): ME = 2.17 × 0.0056 ≈ 0.0122.
5. Find the lower and upper bounds of the confidence interval by subtracting and adding the margin of error to the mean, respectively: Lower bound = 12.05 - 0.0122 ≈ 12.0378, Upper bound = 12.05 + 0.0122 ≈ 12.0622.
So, the 97% confidence interval for this sample is approximately (12.0378, 12.0622), which is closest to the option (12.038, 12.062).
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Suppose you need to pump air into a basketball that is completely deflated. The deflated basketball weighs 0.615 kilograms. After being inflated, the ball weighs 0.618 kilograms. The basketball has a diameter of 0.17 meters. What is the density of air in the ball? Assume the ball is perfectly spherical. Round your answer to two decimal places.
The density of the air inside the ball is approximately 11.69 kg/m³.
What is density?Density is a unit of measurement for mass per volume.
It is calculated by dividing an object's mass by its volume, and is typically denoted by the symbol "."
In the SI system, the unit of density is typically kilogrammes per cubic metre (kg/m3).
To solve this problem, we need to use the equation for the density of an object: density = mass / volume
We can find the volume of the basketball by using the formula for the volume of a sphere: volume = (4/3)πr³
Since the basketball has a diameter of 0.17 meters, its radius is 0.085 meters. When we use this value as a substitute in the volume formula, we get:
volume = (4/3)π(0.085)³ = 0.0002562834 m³
To find the mass of the air inside the ball, we subtract the mass of the deflated ball from the mass of the inflated ball:
mass of air = 0.618 kg - 0.615 kg = 0.003 kg
Now we can calculate the density of the air inside the ball:
density = mass of air / volume = 0.003 kg / 0.0002562834 m³ = 11.69 kg/m³.
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solve the differential equation ( y 13 x ) d y d x = 1 x . (y13x)dydx=1 x. use the initial condition y ( 1 ) = 4 y(1)=4 . express y 14 y14 in terms of x x .
Substituting x = 1, we get:
y''(1) = 16/√3.
To solve the differential equation (y^(1/3)x)dy/dx = 1/x, we need to separate the variables and integrate both sides with respect to their respective variables.
First, we can rewrite the equation as:
dy/y^(1/3) = (dx/x)
Next, we can integrate both sides:
∫dy/y^(1/3) = ∫dx/x
Integrating the left side, we use the substitution u = y^(1/3), du = (1/3)y^(-2/3)dy:
3∫du/u = ln|u| + C1 = ln|y^(1/3)| + C1
Integrating the right side, we get:
∫dx/x = ln|x| + C2
Putting the two integrals together, we have:
ln|y^(1/3)| = ln|x| + C
where C = C2 - C1
To solve for y, we can exponentiate both sides:
|y^(1/3)| = e^C|x|
Since y(1) = 4, we can use this initial condition to solve for the constant C:
|4^(1/3)| = e^C|1|
C = ln(4^(1/3)) = ln(2/√3)
Substituting C into the equation above, we get:
|y^(1/3)| = e^(ln(2/√3))|x| = (2/√3)|x|
Squaring both sides and solving for y, we get:
y = (2/√3)^3x^3 = (8/3√3)x^3
Finally, to express y''(1) in terms of x, we take the second derivative of y:
y = (8/3√3)x^3
y' = 8x^2/√3
y'' = 16x/√3
Substituting x = 1, we get:
y''(1) = 16/√3.
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Use traces to sketch the surface.5x2 − y2 + z2 = 0Identify the surface.o parabolic cylIdentify the surface.parabolic cylinderhyperboloid of one sheethyperboloid of two sheetselliptic paraboloidhyperbolic paraboloidellipsoidelliptic cylinderelliptic cone
The surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] is a parabolic cylinder oriented along the x-axis, and it has a double cone shape in the yz-plane.
To sketch the surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] using traces, we can set two of the variables equal to constants and solve for the third variable.
Setting z = 0, we get [tex]5x^2 - y^2 = 0[/tex], which is the equation of a parabolic cylinder oriented along the x-axis. This means that the surface has a cross-section in the z=0 plane that is a parabola, and the surface extends infinitely in the z-direction.
Setting x = 0, we get [tex]-y^2 + z^2 = 0[/tex], which is the equation of a double cone oriented along the y- and z-axes. This means that the surface has a cross-section in the x=0 plane that is a double hyperbola, and the surface extends infinitely in both the positive and negative x-directions.
Setting y = 0, we get [tex]5x^2 + z^2 = 0[/tex], which is the equation of a single point at the origin (0,0,0).
Therefore, the surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] is a parabolic cylinder oriented along the x-axis, and it has a double cone shape in the yz-plane. This surface is a degenerate quadric surface, meaning that it is not a smooth surface but rather a surface that has been flattened or collapsed in some way. In this case, the surface is a degenerate hyperboloid of one sheet.
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The surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] is a parabolic cylinder oriented along the x-axis, and it has a double cone shape in the yz-plane.
To sketch the surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] using traces, we can set two of the variables equal to constants and solve for the third variable.
Setting z = 0, we get [tex]5x^2 - y^2 = 0[/tex], which is the equation of a parabolic cylinder oriented along the x-axis. This means that the surface has a cross-section in the z=0 plane that is a parabola, and the surface extends infinitely in the z-direction.
Setting x = 0, we get [tex]-y^2 + z^2 = 0[/tex], which is the equation of a double cone oriented along the y- and z-axes. This means that the surface has a cross-section in the x=0 plane that is a double hyperbola, and the surface extends infinitely in both the positive and negative x-directions.
Setting y = 0, we get [tex]5x^2 + z^2 = 0[/tex], which is the equation of a single point at the origin (0,0,0).
Therefore, the surface defined by [tex]5x^2 - y^2 + z^2 = 0[/tex] is a parabolic cylinder oriented along the x-axis, and it has a double cone shape in the yz-plane. This surface is a degenerate quadric surface, meaning that it is not a smooth surface but rather a surface that has been flattened or collapsed in some way. In this case, the surface is a degenerate hyperboloid of one sheet.
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A figure that 8 inc long, 7 inch wide and 6 inch tall
The volume of the figure is 336 cubic inches.
To find the volume of the figure, we simply multiply its width, height, and length together using the formula V = l x w x h.
We have been Given that the width is 7 inches exactly, the height is given as 6 inches exactly, and the length is 8 inches, we can substitute all of these values into the basic formula of volume:
V = l x w x h
V = 8 x 7 x 6
V = 336
Therefore, the volume of the figure is 336 cubic inches.
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Complete Question:
What is the volume of a figure that is 7 inches wide, 6 inches tall and 8 inches long?
Change from rectangular to cylindrical coordinates. (Let r ≥ 0 and 0 ≤ θ ≤ 2π.) (a) (−9, 9, 9) (b) (−8, 8 3 , 5)
(a) Cylindrical coordinates for (a): (r, θ, z) = (√162, 3π/4, 9)
(b) Cylindrical coordinates for (b): (r, θ, z) = (√256, 4π/3, 5)
Cylindrical coordinates can be defined as three sets of coordinates used to locate a point in a cylindrical coordinate system. In two dimensions, the position of a point can be expressed in Cartesian and polar coordinates. When polar coordinates are extended to the 3D plane, an additional coordinate is added. Together, these three measurements form cylindrical coordinates. Coordinates define both distance and angle.
The radial distance, azimuth, and height of the plane from a point are expressed in cylindrical coordinates. Cylindrical-coordinate systems can be used to describe systems with rotational symmetry.
To convert from rectangular coordinates to cylindrical coordinates, we use the following equations:
r = sqrt(x^2 + y^2)
theta (θ) = arctan(y/x)
z = z
For part (a), we have the rectangular coordinates (-9, 9, 9). Using the above equations, we get:
r = sqrt((-9)^2 + 9^2) = 9 sqrt(2)
theta = arctan(9/-9) = -π/4 (since the point is in the third quadrant)
z = 9
So the cylindrical coordinates for part (a) are (9 sqrt(2), -π/4, 9).
For part (b), we have the rectangular coordinates (-8, 8 sqrt(3), 5). Using the above equations, we get:
r = sqrt((-8)^2 + (8 sqrt(3))^2) = 16
theta = arctan(8 sqrt(3)/-8) = -π/3 (since the point is in the third quadrant)
z = 5
So the cylindrical coordinate for part (b) is (16, -π/3, 5).
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DUE FRIDAY PLEASE HELP WELL WRITTEN ANSWERS ONLY!!!!
A company that produces television shows is interested in what type of show people would like to watch for a prime time slot (crime drama, animated comedy, or reality contest). The company asks, “Which show would you be most likely to watch during prime time?
• Mr. Winslow
• Kibble
• Extreme Mountain Hunter: a show in which 20 contestants attempt to climb some of the tallest mountains in the world using only equipment they create from nature
Will this question likely produce data that would allow the company to answer the question they are interested in? Explain your reasoning.
Answer:
it will not because only "extreme mountain hunter" is given a vivid description. potential viewers might not know what the first two are about, therefore skewing the results.
Step-by-step explanation:
Which is a table of values for y=x-6
This set represents all possible values of [tex]y[/tex] that result from substituting each value of [tex]x[/tex] in the given domain into the equation [tex]y=x-6.[/tex]
What is the intercept?The equation [tex]y=x-6[/tex] is in slope-intercept form, where the slope is 1 and the y-intercept is [tex]-6[/tex]. This means that for any value of x, the corresponding value of y can be found by subtracting 6 from x.
Here are some values for y, given different values of x:
When [tex]x=0, y=(-6)[/tex]
When [tex]x=1, y=(-5)[/tex]
When [tex]x=2, y=(-4)[/tex]
When [tex]x=3, y=(-3)[/tex]
When [tex]x=4, y=(-2)[/tex]
When [tex]x=5, y=(-1)[/tex]
When [tex]x=6, y=0[/tex]
When [tex]x=7, y=1[/tex]
Therefore, This set represents all possible values of y that result from substituting each value of x in the given domain into the equation [tex]y=x-6.[/tex]
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a certain surgical procedure is successful only 40% of the time.a) what is the probability that exactly 7 of 11 surgeries are successful?
The probability of exactly 7 successful surgeries out of 11 is 0.168.
This problem can be solved using the binomial probability formula:
[tex]P(X = k) = C(n, k)[/tex] ×[tex]p^k[/tex] × [tex](1-p)^{n-k}[/tex]
where:
P(X = k) is the probability of getting exactly k successes
n is the number of trials (in this case, 11 surgeries)
k is the number of successful surgeries
p is the probability of success in a single trial (in this case, 0.4)
C(n, k) is the number of ways to choose k successes from n trials, which is calculated as n choose [tex]k = n! / (k![/tex]×[tex](n-k)!)[/tex]
Using this formula, we can plug in the values and calculate the probability of getting exactly 7 successful surgeries:
[tex]P(X = 7) = C(11, 7)[/tex] × [tex]0.4^7[/tex]× [tex]0.6^{11-7}[/tex]
[tex]= 330[/tex] × [tex]0.0390625[/tex] × [tex]0.279936[/tex]
[tex]= 0.0968[/tex] (rounded to four decimal places)
This situation can be modeled by a binomial distribution with n = 11 surgeries and p = 0.4 probability of success.
The probability of exactly k successes out of n trials is given by the binomial probability formula:
P(k successes) = (n choose k) × p^k × (1 - p)^(n - k)
where (n choose k) is the binomial coefficient, which represents the number of ways to choose k items from a set of n items.
Using this formula, we can calculate the probability of exactly 7 successful surgeries out of 11 as follows:
[tex]P(7 successes) = (11 choose 7)[/tex] × [tex]0.4^7[/tex]× [tex]0.6^4[/tex]
[tex]= (11! / (7![/tex] × [tex]4!))[/tex] × [tex]0.4^7[/tex] × [tex]0.6^4[/tex]
[tex]= 330[/tex] × [tex]0.004096[/tex] × [tex]0.1296[/tex]
[tex]= 0.168[/tex]
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using the definition of the dual of a problem in standardform, find the dual of the linear programmingproblem maximize z = ctx dtx' subjectto ax bx' < b x > 0, x' unrestricted
The dual of the given linear programming problem in standard form is:
Minimize w = b^T y
Subject to: a^T y + b^T y' ≥ ct
y ≥ 0
y' unrestricted.
To find the dual of a linear programming problem in standard form, we follow these steps:
1. Write the primal problem in standard form:
Maximize z = c^T x
Subject to: Ax ≤ b
x ≥ 0
where x is a vector of decision variables, c is a vector of coefficients for the objective function, A is a matrix of coefficients for the constraints, and b is a vector of constants for the constraints.
2. Write the dual problem in standard form:
Minimize w = b^T y
Subject to: A^T y ≥ c
y ≥ 0
where y is a vector of dual variables, b is a vector of constants for the primal constraints, and A^T is the transpose of matrix A.
Applying this process to the given linear programming problem, we get:
Primal problem:
Maximize z = c^T x
Subject to: Ax ≤ b
x ≥ 0
where c = ct and x' = x
Maximize z = ct x
Subject to: ax ≤ b
bx ≤ d
x ≥ 0
x' unrestricted
Dual problem:
Minimize w = b^T y
Subject to: A^T y ≥ c
y ≥ 0
where b = (b, d) and A^T = (a, b)
Minimize w = b^T y
Subject to: a^T y + b^T y' ≥ ct
y ≥ 0
y' unrestricted
Therefore, the dual of the given linear programming problem in standard form is:
Minimize w = b^T y
Subject to: a^T y + b^T y' ≥ ct
y ≥ 0
y' unrestricted.
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Prepare an income statement for Hansen Realty for the year ended December 31, 2017. Beginning inventory was $1,245. Ending inventory was $1,597. (Input all amounts as positive values.)
Sales $ 34,600
Sales returns and allowances 1,089
Sales discount 1,149
Purchases 10,362
Purchase discounts 537
Depreciation expense 112
Salary expense 5,050
Insurance expense 2,450
Utilities expense 207
Plumbing expense 247
Rent expense 177
HANSEN REALTY
Income Statement
For Year Ended December 31, 2017
(Click to select)DepreciationCost of merchandise (goods) soldPurchasesSalaryRentInsuranceUtilitiesPlumbingPurchase discountsNet sales $
(Click to select)DepreciationPurchasesSalaryInsurancePlumbingRentUtilitiesCost of merchandise (goods) soldPurchase discountsNet sales (Click to select)Gross profit from salesGross loss from sales $
Operating expenses: (Click to select)Net salesRentCost of merchandise (goods) soldInsurancePlumbingDepreciationPurchasesSalaryUtilitiesPurchase discounts $ (Click to select)PlumbingInsuranceRentNet salesCost of merchandise (goods) soldPurchasesUtilitiesDepreciationPurchase discountsSalary (Click to select)SalaryRentNet salesUtilitiesPurchase discountsPurchasesPlumbingDepreciationInsuranceCost of merchandise (goods) sold (Click to select)SalaryRentPlumbingDepreciationPurchasesInsuranceCost of merchandise (goods) soldUtilitiesNet salesPurchase discounts (Click to select)Net salesUtilitiesCost of merchandise (goods) soldDepreciationPurchase discountsInsuranceSalaryPlumbingRentPurchases (Click to select)UtilitiesCost of merchandise (goods) soldPurchasesInsurancePlumbingPurchase discountsDepreciationNet salesRentSalary Total operating expenses (Click to select)Net incomeNet loss $
Operating Expenses: $8,243
Net Income: $15,891
HANSEN REALTY
Income Statement
For Year Ended December 31, 2017
Net Sales: $34,600 - $1,089 - $1,149 = $32,362
Cost of Goods Sold:
Beginning Inventory: $1,245
Purchases: $10,362 - $537 = $9,825
Total Cost of Merchandise Available for Sale: $11,070
Ending Inventory: $1,597
Cost of Goods Sold: $11,070 - $1,597 - $1,245 = $8,228
Gross Profit: $32,362 - $8,228 = $24,134
Operating Expenses:
Depreciation Expense: $112
Salary Expense: $5,050
Insurance Expense: $2,450
Utilities Expense: $207
Plumbing Expense: $247
Rent Expense: $177
Total Operating Expenses: $8,243
Net Income: $24,134 - $8,243 = $15,891
Therefore, the Income Statement for Hansen Realty for the year ended December 31, 2017 is as follows:
HANSEN REALTY
Income Statement
For Year Ended December 31, 2017
Net Sales: $32,362
Cost of Goods Sold: $8,228
Gross Profit: $24,134
Operating Expenses: $8,243
Net Income: $15,891
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como simplificar 1- 4/9
Answer:
[tex]=\frac{5}{9}[/tex]
Step-by-step explanation:
[tex]=\frac{9}{9}-\frac{4}{9}[/tex]
then
[tex]=\frac{9-4}{9}[/tex]
[tex]\mathrm{Subtract\:the\:numbers:}[/tex]
[tex]=\frac{5}{9}[/tex]
Answer:
[tex]=\frac{5}{9}[/tex]
Step-by-step explanation:
[tex]=\frac{9}{9}-\frac{4}{9}[/tex]
then
[tex]=\frac{9-4}{9}[/tex]
[tex]\mathrm{Subtract\:the\:numbers:}[/tex]
[tex]=\frac{5}{9}[/tex]
find the general solution of the given system. dx dt = − 5 2 x 4y dy dt = 3 4 x − 3y
The general solution of the given system is: x(t) = c1 e^(-5/2t), y(t) = c2 c3 e^(-4/3c2t), where c1, c2, and c3 are arbitrary constants.
To find the general solution of the given system, we can use the method of separation of variables.
First, we rewrite the system in the form:
dx/dt = -5/2 x + 0 y
dy/dt = 3/4 x - 3y
Then, we separate the variables by putting all the x terms on one side and all the y terms on the other side:
dx/dt + (5/2)x = 0
dy/dt + 3y = (3/4)x
Next, we solve each equation separately. For the first equation, we have:
dx/dt + (5/2)x = 0
This is a first-order linear homogeneous differential equation, which has the general solution:
x(t) = c1 e^(-5/2t)
where c1 is an arbitrary constant.
For the second equation, we have:
dy/dt + 3y = (3/4)x
This is a first-order linear non-homogeneous differential equation, which has a particular solution of the form:
y(t) = c2 x(t)
where c2 is another arbitrary constant.
To find the general solution, we combine the two solutions we found for x and y:
x(t) = c1 e^(-5/2t)
y(t) = c2 x(t)
Substituting y(t) into the second equation, we get:
dy/dt + 3y = (3/4)x
c2 dx/dt + 3c2 x = (3/4)x
dx/dt + (4/3)c2 x = 0
This is another first-order linear homogeneous differential equation, which has the general solution:
x(t) = c3 e^(-4/3c2t)
where c3 is another arbitrary constant.
Finally, we substitute this solution for x back into the equation for y to get:
y(t) = c2 x(t) = c2 c3 e^(-4/3c2t)
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Change from rectangular to cylindrical coordinates. (Let r>=0 and 0<=σ<=2π.)
(a) (-8, 8, 8)
(b) (4, 3 , 9)
To change from rectangular to cylindrical coordinates for points (a) (-8, 8, 8) and (b) (4, 3, 9):
(a) In cylindrical coordinates, the point (-8, 8, 8) is (r, σ, z) = (√128, 3π/4, 8).
(b) For the point (4, 3, 9), the cylindrical coordinates are (r, σ, z) = (5, 0.93, 9).
To convert from rectangular (x, y, z) to cylindrical (r, σ, z) coordinates, follow these steps:
1. Calculate r: r = √(x² + y²)
2. Calculate σ: σ =cylindrical coordinates(y/x) (note that σ is between 0 and 2π)
3. Keep the same z value.
For point (a):
1. r = √((-8)² + 8²) = √128
2. σ = arctan(8/-8) = arctan(-1) = 3π/4 (adjusted to be in the range 0 to 2π)
3. z = 8
For point (b):
1. r = √(4² + 3²) = 5
2. σ = arctan(3/4) ≈ 0.93 (adjusted to be in the range 0 to 2π)
3. z = 9
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The Height of Harrys tower is 45/100 of a meter and a height of Jenny’s tower is 55/100 what is the difference in the Heights
Answer:
10m
Step-by-step explanation:
Difference means minus (-)
55/100 - 45/100
lcm=100
multiply through by 100
100×55/100 -100×45/100
=55 - 45
=10m