Find the area of each triangle. Round intermediate values to the nearest 10th. use the rounded value to calculate the next value. Round your final answer to the nearest 10th.

The answer is (a) 0.50.

To find the percent of observations that fall between 1 and 4, we need to find the area under the curve between 1 and 4 and divide by the total area under the curve.

The total area under the curve is equal to 1, since this is a probability density function.

The area under the curve between 1 and 4 can be found by calculating the integral of the density function from x = 1 to x = 4:

∫1^4 (1/6) dx = (1/6)(4-1) = 1/2

So the area between 1 and 4 is 1/2, and the percent of observations that fall in this range is:

(1/2) * 100% = 50%

Therefore, the answer is (a) 0.50.

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The possible rational zeros of the polynomial are of the form:

±1, ±3, ±7, ±21, ±1/3, ±7/3.

The Rational Root Theorem proclaims that if any polynomial is composed of integer coefficients, then its potential rational zeros will have a numerator which is a divisor of the constant term and a denominator that can be factored into the leading coefficient.

Considering the case at hand, with the leading coefficient as 3 and the constant term being -21, the reasonable zeros conceived are in this format:

±1, ±3, ±7, ±21, ± 1/3, ± 7/3.

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We have derived the desired expressions: E[Y] = μE[N], Var(Y) = σ²E[N] + μ²Var[N]

To answer your question, let's start by defining the given terms:

N = number of shops visited (positive integer random variable)
Xi = amount of money spent in the i-th shop
Y = X1 + X2 + ... + XN (total amount of money spent)
μ = mean of the Xi's
σ² = variance of the Xi's

We are given that N and all Xi are independent, and we want to show that:

E[Y] = μE[N]
Var(Y) = σ²E[N] + μ²Var[N]

First, let's find E[Y]:

E[Y] = E[X1 + X2 + ... + XN]
By linearity of expectation:
E[Y] = E[X1] + E[X2] + ... + E[XN]

Since all Xi's have the same mean (μ), we can rewrite this as:

E[Y] = Nμ
Since N is a random variable, we take the expectation of N:
E[Y] = μE[N]

Next, let's find Var(Y):

Var(Y) = Var(X1 + X2 + ... + XN)

As N and all Xi are independent, we can say:

Var(Y) = Var(X1) + Var(X2) + ... + Var(XN)

Since all Xi's have the same variance (σ²), we can rewrite this as:

Var(Y) = Nσ²

Now, we need to find the variance of N:

Var(Y) = σ²E[N] + μ²Var[N]

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The least distance between P and (50,0) is 25sqrt(10) units.

The line y=3x can be expressed in the form y=mx+b, where m is the slope and b is the y-intercept. In this case, m=3 and b=0, so the equation of the line can be written as y=3x.

Let P=(x,y) be the point on the line y=3x that is closest to the point (50,0). Then the vector from P to (50,0) is orthogonal to the line y=3x. The direction vector of the line y=3x is (1,3), so the direction vector of the vector from P to (50,0) is (-3,1).

Let Q=(50,0) and R=(x,y) be the two points. The vector from Q to R is given by:

v = R - Q = (x-50, y-0) = (x-50, y)

Since the vector v is orthogonal to the direction vector (-3,1) of the line y=3x, their dot product is zero:

v . (-3,1) = 0

Substituting v and the equation y=3x, we get:

(x-50, 3x) . (-3,1) = 0

Expanding the dot product, we get:

-3(x-50) + 3x = 0

Simplifying the equation, we get:

x = 25

Substituting x=25 into the equation y=3x, we get:

y = 75

Therefore, the point P on the line y=3x that is closest to the point (50,0) is (25,75), and the least distance between P and (50,0) is the length of the vector v:

|v| = sqrt((x-50)^2 + y^2) = sqrt((25-50)^2 + 75^2) = sqrt(6250) = 25sqrt(10)

So the least distance between P and (50,0) is 25sqrt(10) units.

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The solutions of the equation sec² x - 4 = 0 are x = π/3 + 2kπ, 2π/3 + 2kπ, 4π/3 + 2kπ, 5π/3 + 2kπ. Therefore, option d. is correct.

We start by solving for x in the equation sec² x - 4 = 0:

sec² x - 4 = 0
sec² x = 4
sec x = ±2

Recall that sec x = 1/cos x, so we can rewrite the equation as:

1/cos² x = 4
cos² x = 1/4
cos x = ±1/2

Now we need to find all values of x that satisfy cos x = ±1/2. These values are:

x = π/3 + 2kπ or x = 5π/3 + 2kπ (for cos x = 1/2)
x = 2π/3 + 2kπ or x = 4π/3 + 2kπ (for cos x = -1/2)

Combining these solutions, we get:
x = π/3 + 2kπ, 2π/3 + 2kπ, 4π/3 + 2kπ, 5π/3 + 2kπ

Therefore, the correct answer is d. pi/3 + 2k pi, 2pi/3 + 2k pi, 4pi/3 + 2k pi, 5pi/3 + 2k pi.

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Answer:

Step-by-step explanation:

To compare which option is the better deal, we need to determine the final price of the sofa after applying each discount.

For Choice A, the price after the 15% discount is:

1250 - (0.15 x 1250) = 1250 - 187.50 = $1062.50

For Choice B, the price after the $200 rebate is:

1250 - 200 = $1050

Therefore, Choice B, with a $200 rebate, is the better deal for the customer as it results in a lower final price of $1050, compared to Choice A, which results in a final price of $1062.50 after the 15% discount.

Step-by-step explanation:

Let's represent the cost of one peach as "x" and the cost of one nectarine as "y". We can set up a system of equations using the given information:

4x + 12y = 2.28

2x + 14y = 2.10

We can write this system in matrix form as AX = B, where

A = |4 12|

|2 14|

X = |x|

|y|

B = |2.28|

|2.10|

To solve for X, we can use the formula X = A^(-1)B, where A^(-1) is the inverse of matrix A.

First, we need to find the inverse of matrix A:

|4 12| |7/30 -2/15|

|2 14| = |-1/30 2/15|

Now we can use the formula X = A^(-1)B:

|x| |7/30 -2/15||2.28| |0.24|

|y| = |-1/30 2/15||2.10| = |0.12|

Therefore, the cost for a peach is $0.24 (rounded to two decimal places) and the cost for a nectarine is $0.12 (rounded to two decimal places).

The distribution of (M,N) is a bivariate Poisson distribution with parameters λ_1 = 9 (rate of local calls) and λ_2 = 3 (rate of long distance calls) since the rates of the two processes are independent Poisson processes.  and the number of people on line are 60.

The joint probability mass function of (M,N) is given by:

[tex]P(M=m, N=n) = e^{-(\lambda_1 +\lambda_2) (lambda_1^m/ m!) (\lambda_2^n/ n!)[/tex]

The number of people on the line is the expected value of the total number of minutes of all calls, which can be calculated as E[10M + 5N] = 10E[M] + 5E[N].

Since M and N are independent Poisson random variables, we have E[M] = λ_1/(1-p_1) and E[N] = λ_2/(1-p_2), where p_1 and p_2 are the probabilities of a call being local or long distance, respectively.

Substituting in the given values, we get:

E[10M + 5N] = 10(9/3) + 5(3/1) = 45 + 15 = 60

Therefore, the expected number of people on the line is 60.

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a= 1

h= 1

k= 0

The graph moves to the right 1 unit.

The bias of mu1 is 0, the variance of mu1 is 1/2, and the mean squared error of mu1 is 1/2.

We need to find the bias, variance, and mean squared error of mu1, given that x1 and x2 are independent with an unknown mean (mu) and a known variance (sigma^2 = 1), and mu1 = (x1 + x2)/2.

Step 1: Compute the expected value of mu1.
E(mu1) = E((x1 + x2)/2) = (E(x1) + E(x2))/2 = (mu + mu)/2 = mu

Step 2: Calculate the bias of mu1.
Bias (mu1) = E (mu1) - mu = mu - mu = 0

Step 3: Calculate the variance of mu1.
Var(mu1) = Var((x1 + x2)/2) = (1/4) * (Var(x1) + Var(x2)) = (1/4) * (1 + 1) = 1/2

Step 4: Calculate the mean squared error of mu1.
MSE(mu1) = Bias(mu1)^2 + Var(mu1) = 0^2 + 1/2 = 1/2

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Mr. David Miller, a statistician, is interested in selecting a sample. He want to give equal probability to each population unit to be selected in the sample. This method is called option (a) simple random sampling

Simple random sampling is a method of selecting a sample from a population where each unit has an equal chance of being selected. This means that every member of the population has the same probability of being chosen for the sample, making it a fair and unbiased method of sampling.

The process involves selecting units at random, without any specific order or pattern. Simple random sampling is commonly used in research studies to obtain representative samples from a population, as it ensures that the sample is a true reflection of the population and reduces the risk of bias or error in the findings.

Therefore, the correct option is (a) simple random sampling

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Therefore, there will be approximately 20,480 bacteria after 30 minutes. The correct answer is (b) 20,480 bacteria.

In this experiment, we have an initial bacterial count of 5,000 at time 0 and a count of 8,000 at 10 minutes. Since the growth is exponential, we can use the exponential growth formula:

[tex]N(t) = N₀ * (1 + r)^t[/tex]

Where N(t) is the number of bacteria at time t, N₀ is the initial number of bacteria, r is the growth rate, and t is the time.

First, let's find the growth rate using the data provided:

[tex]8000 = 5000 * (1 + r)^{10\\\\(1 + r)^{10} = 8000 / 5000 = 1.6[/tex]

Now, let's find the 10th root of 1.6 to find the growth rate (1 + r):

[tex]1 + r = 1.6^{1/10} =1.0481[/tex]

So, the growth rate (r) is approximately 0.0481.

Next, we want to find the number of bacteria after 30 minutes:

N(30) = 5000 * (1.0481)^30 ≈ 20480.18

Therefore, there will be approximately 20,480 bacteria after 30 minutes. The correct answer is (b) 20,480 bacteria.

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The missing values of the polynomial are listed as below

The Polynomial x² - 8x + 15 is factored to (x - 3)(x - 5)

comparing with (ax + b)(cx + d)

a = 1, b  = -3, c = 1 d = -5.

the Polynomial 2x² - 24x is factored to 2x(ax + b)(cx + d)

2x² - 24x = 2x(x + 0)(2x - 6) = 2x(x)(2x - 6)

comparing with 2x(ax + b)(cx + d)

a = 1, b  = 0, c = 2 d = -6.

2x² - 24x = 2x(x)(2x - 6)

the Polynomial 6x² + 14x + 4 is factored to (2x + 2) (3x + 2)

comparing with  (ax + b)(cx + d)

a = 2, b = 2, c = 3, and d = 2.

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The missing values of the polynomial are listed as below

The Polynomial x² - 8x + 15 is factored to (x - 3)(x - 5)

comparing with (ax + b)(cx + d)

a = 1, b  = -3, c = 1 d = -5.

the Polynomial 2x² - 24x is factored to 2x(ax + b)(cx + d)

2x² - 24x = 2x(x + 0)(2x - 6) = 2x(x)(2x - 6)

comparing with 2x(ax + b)(cx + d)

a = 1, b  = 0, c = 2 d = -6.

2x² - 24x = 2x(x)(2x - 6)

the Polynomial 6x² + 14x + 4 is factored to (2x + 2) (3x + 2)

comparing with  (ax + b)(cx + d)

a = 2, b = 2, c = 3, and d = 2.

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The system of inequalities that can be used to determine the number of Tyler classes, X, and the number of preschool classes, Y, is:

8X + 12Y ≤ 96

X + Y ≤ 10

In mathematics, a system of inequalities refers to a set of two or more inequalities involving one or more variables. The solution to a system of inequalities is a set of values for the variables that satisfy all the inequalities in the system simultaneously.

For example, a system of two linear inequalities in two variables x and y can be written in the form:

ax + by ≤ c

dx + ey ≤ f

where a, b, c, d, e, and f are constants. The solution to this system is the set of all points (x,y) that satisfy both inequalities simultaneously.

Let X be the number of Tyler classes and Y be the number of preschool classes.

The number of students in a Tyler class is 8, so the total number of Tyler class students is 8X. Similarly, the number of students in a preschool class is 12, so the total number of preschool students is 12Y.

The total number of students cannot exceed 96, so we have the inequality:

8X + 12Y ≤ 96

The daycare center has up to 10 rooms available, so we have the inequality:

X + Y ≤ 10

Therefore, the system of inequalities that can be used to determine the number of Tyler classes, X, and the number of preschool classes, Y, is:

8X + 12Y ≤ 96

X + Y ≤ 10

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a) The probability that daily sales will fall between 2,500 and 3,000 gallons is 0.2.

b) The probability that the service station will sell at least 4,000 gallons is 0.3.

c) The probability that the station will sell exactly 2,500 gallons is 0.1.

Gasoline sold daily is uniformly distributed between 2,000 and 5,000 gallons.

a) To find the probability that daily sales will fall between 2,500 and 3,000 gallons, we need to find the area under the uniform distribution curve between 2,500 and 3,000.

Since the distribution is uniform, the probability density function (PDF) is a constant:

f(x) = 1 / (5000 - 2000) = 1 / 3000, for 2000 <= x <= 5000.

The probability of the daily sales being between 2,500 and 3,000 gallons is equal to the area under the PDF curve between 2,500 and 3,000:

Therefore, the probability that daily sales will fall between 2,500 and 3,000 gallons is 0.1667 or 16.67%.

b) To find the probability that the service station will sell at least 4,000 gallons,

we need to find the area under the uniform distribution curve from 4,000 to 5,000 gallons:

Therefore, the probability that the service station will sell at least 4,000 gallons is 0.3333 or 33.33%.

c) To find the probability that the station will sell exactly 2,500 gallons, we need to find the area under the uniform distribution curve at 2,500 gallons:

P(X = 2500) = 0 (since the probability of a single point is zero for a continuous distribution)

Therefore, the probability that the station will sell exactly 2,500 gallons is zero.

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The comparison test with the series ∑ 1 / n^(9/2) shows that the series converges.

To use the comparison test to determine if the series ∑ (sin^2 n) / n^(9/2) converges or diverges, we need to find a known series with terms that are greater than or equal to the terms of the given series. If the known series converges, then the given series also converges. If the known series diverges, then we cannot conclude anything about the convergence of the given series.

We can use the comparison test with the series ∑ [tex]1 / n^(9/2)[/tex] to determine the convergence or divergence of the given series. Since [tex]sin^2[/tex] n is always between 0 and 1, we have:

0 ≤ [tex](sin^2 n) / n^(9/2)[/tex] ≤ [tex]1 / n^(9/2)[/tex]

Therefore, by the comparison test, if the series ∑ [tex]1 / n^(9/2)[/tex] converges, then the series ∑ [tex](sin^2 n) / n^(9/2)[/tex] also converges. Conversely, if the series ∑ [tex]1 / n^(9/2)[/tex] diverges, then the series ∑ [tex](sin^2 n) / n^(9/2)[/tex] also diverges.

The series ∑ [tex]1 / n^(9/2)[/tex] is a p-series with p = 9/2 > 1, so it converges. Therefore, by the comparison test, the given series ∑ [tex](sin^2 n) / n^(9/2)[/tex] also converges.

The correct answer is B. The comparison test with the series ∑ [tex]1 / n^(9/2)[/tex] shows that the series converges.

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The series Σ^[infinity]_n=1 (3 / (4^n + 5)) converges.

To determine whether the series converges or diverges, consider the series Σ^[infinity]_n=1 (3 / (4^n + 5)).

Step 1: Apply the Comparison Test. We will compare our given series with a known convergent or divergent series. In this case, let's compare it to the geometric series Σ^[infinity]_n=1 (3 / 4^n).

Step 2: Since 4^n + 5 > 4^n, we have (3 / (4^n + 5)) < (3 / 4^n). Now we need to determine if the geometric series converges.

Step 3: Check the geometric series. A geometric series converges if the absolute value of the common ratio is less than 1. The common ratio in our case is 1/4 (since each term is multiplied by 1/4 to get the next term). The absolute value of the common ratio |1/4| = 1/4, which is less than 1.

Step 4: Conclude the convergence of the geometric series. Since the geometric series Σ^[infinity]_n=1 (3 / 4^n) converges, and our given series is term-by-term smaller, it also converges by the Comparison Test.

Thus, the series Σ^[infinity]_n=1 (3 / (4^n + 5)) converges.

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Answer:.40 or .4 would be the correct answer

Answer:

0.4

Step-by-step explanation:

40% can be written as 0.4. Recall that to get a decimal to a percent or vice versa, you can move the decimal place 2 places to the right or left to get the desired outcome. In this case, the decimal needs to move 2 places to the left leaving you with the decimal 0.4 or 4 parts out of 10 (40 parts out of 100= 40%)

The following can be answered by the concept of Trigonometry.

The rearranged functions for each case, using appropriate mathematical techniques and identities:

(a) For 1 - cos(x) / x², we can use the identity sin²(x) + cos²(x) = 1 and rearrange as follows:
1 - cos(x) = sin²(x) / (1 + cos(x))
Then, divide by x² to get the rearranged function:
(sin²(x) / (1 + cos(x))) / x²

(b) For log(x + 1) - log(x) with x being large, we can use the logarithmic property of subtraction to combine the logs:
log((x + 1) / x)
As x is large, the expression simplifies to:
log(1 + (1/x))

(c) For sin(a + x) - sin(a), we can use the angle sum identity for sine:
sin(a + x) - sin(a) = (sin(a)cos(x) + cos(a)sin(x)) - sin(a)
Rearrange to get:
cos(a)sin(x) + sin(a)(cos(x) - 1)

(d) For √(1 + x - 1), we can simplify by removing the 1 and -1 inside the square root:
√(x)

(e) For √(4 + x - 2) / x, we can first simplify inside the square root:
√(2 + x) / x

These rearranged functions help avoid loss-of-significance errors in each of the given cases.

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The total number of n-digit numbers with all digits odd, such that 1 and 3 each occur a nonzero, even number of times is  5^n * (n+1)^3 + 2 * 5^(n-1) * n * (n-2)^2.

Let's first consider the odd digits that can appear in an n-digit number, which are 1, 3, 5, 7, and 9. Since each digit can be used independently of the others, the total number of possible n-digit numbers with all digits odd is simply 5^n.

Now let's consider the condition that 1 and 3 each occur a nonzero, even number of times. There are two possibilities either both digits occur an even number of times (0, 2, 4, 6, or 8), or both digits occur an odd number of times (1, 3, 5, 7, or 9). We will consider each case separately.

Both 1 and 3 occur an even number of times.

In this case, we can choose the number of times each of the other three digits appears independently of each other.

Each of the other three digits can appear 0, 1, 2, 3, ..., or n times. Thus, the total number of n-digit numbers with all digits odd and with both 1 and 3 appearing an even number of times is

5^n * (n+1)^3

This is because there are (n+1) choices for each of the three remaining odd digits (since each can appear 0, 1, 2, ..., or n times), and we have 3 remaining digits to choose.

Both 1 and 3 occur an odd number of times.

In this case, we must choose one of the digits (1 or 3) to appear once, and the other to appear three times. We can choose which digit appears once in 2 ways. We then have (n-1) remaining digits to choose, and we must choose one of the remaining digits to appear 1, 3, 5, ..., or (n-1) times.

The other two remaining digits can appear 0, 2, 4, ..., or (n-2) times each. Thus, the total number of n-digit numbers with all digits odd and with both 1 and 3 appearing an odd number of times is

2 * 5^(n-1) * n * (n-2)^2

This is because we have 2 choices for which digit appears once, 5 choices for the once-appearing digit, n-1 choices for the location of the once-appearing digit, n choices for the digit that appears three times, and (n-2) choices for each of the other two digits (since each can appear 0, 2, 4, ..., or (n-2) times).

Thus, the total number of n-digit numbers is

5^n * (n+1)^3 + 2 * 5^(n-1) * n * (n-2)^2

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Answer:

The answer is trapezoid.

Trapezoids are quadrilaterals that have one set of parallel sides. Some authors claim it is exactly one set of parallel sides, other authors say at least one set of parallel sides, and by doing so allow parallelograms to be a special subgroup of trapezoids.

Trapezoids are not studied in US geometry to anywhere near the same depth as parallelograms. As a result in most US Geometry textbooks the only “special” trapezoids are isosceles trapezoids, where the two sides between the parallel sides are congruent to each other and the right trapezoid, where one of the non-parallel sides is perpendicular to the two parallel sides.

Types of trapezoids:

1. An isosceles trapezoid is a trapezoid where the sides (legs) have the same length, and the base angles have the same measure. It has reflection symmetry. This is possible for acute or right trapezoids. This is a cyclic quadrilateral. You can a draw a circumcircle around such a trapezoid.

2. It is called a right trapezoid (also called right-angled trapezoid) has two adjacent right angles. Right trapezoids are used in the trapezoidal rule for estimating areas under a curve.

3. An acute trapezoid has two adjacent acute angles on its longer base edge. The two adjacent angles on the shorter base edge are obtuse.

4. An obtuse trapezoid has one acute and one obtuse angle on each base.

5. A parallelogram is a trapezoid with two pairs of parallel sides. A parallelogram has central 2-fold rotational symmetry (or point reflection symmetry). It is possible for obtuse or right trapezoids.

6. A tangential trapezoid is a trapezoid that has an incircle.

Yes, the extreme value theorem guarantees the existence of an absolute maximum and minimum for f on the interval [−5,−1] since f is continuous on the closed and bounded interval.

The Extreme Value Theorem states that there is an existence of an absolute maximum and minimum for a function f(x) on a closed interval [a, b] if the function is continuous on that interval. In this case, f(x) = ln(x^2) is continuous on the interval [-5, -1], since x^2 is always non-negative and ln(x^2) is defined for all x in the interval.

Thus, the extreme value theorem applies on the function f(x)=ln(x2) over the interval [−5,−1]. And so, there must be the existence of an absolute maximum and minimum for the function f where f(x)=ln(x2) on the interval [-5, -1].

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The vertex form of the equation is [tex]y = (x-1)^2+2[/tex].

What is vertex form?

when (h,k) is vertex then the straight line  equation is y=[tex]a(x-h)^2+k[/tex].

Here the given straight line vertex is (1,2).

Now applying into equation then,

=> y=a[tex](x-1)^2+2[/tex]

The given line passes through the point (0,2) then,

=> [tex]2=a(0-1)^2+1[/tex]

=> 2=a(1)+1

=> 2=a+1

=> a= 2-1= 1

Then vertex form of the equation is [tex]y = (x-1)^2+2[/tex].

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To the left of z₁ = -1.74 and to the right of z₂ = 1.74, the total area under the standard normal curve is roughly 0.0818.

The standard normal distribution table calculates the probability that a regularly distributed random variable Z, with a mean of 0 and a difference of 1, is not exactly or equal to z. A persistent probability distribution is the normal distribution. It is also known as the Gaussian distribution. It only applies to positive z estimations.

The total area under the standard normal curve to the left of z₁ and to the right of z₂ is the sum of the area to the left of z₁ and the area to the right of z₂.

Using a standard normal distribution table or calculator, we can find:

Therefore, the total area under the standard normal curve to the left of z₁ and to the right of z₂ is:

0.0409 + 0.0409 = 0.0818

Rounding this answer to four decimal places, we get:

0.0818

Therefore, the total area under the standard normal curve to the left of z₁ = -1.74 and to the right of z₂ = 1.74 is approximately 0.0818.

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The charge enclosed by the cube is ±96 times the permittivity of free space ε0.

Gauss's Law states that the electric flux through any closed surface is proportional to the total charge enclosed by the surface. We can use this law to find the charge enclosed by the given cube by constructing a closed surface that encloses the cube and calculating the electric flux through that surface.

The cube has vertices at (±2, ±2, ±2), so we can choose a cube of edge length 4 centered at the origin as our closed surface. The surface is then defined by the six faces of the cube, each of which is a square of area 4² = 16.

The electric field is given by E(x, y, z) = xi + yj + zk, so the flux through each face is the dot product of the field and the outward normal vector to the face, multiplied by the area of the face:

Φ = ∫ E · dA = ∫ (xi + yj + zk) · (±1, 0, 0) dA

= ± ∫ x dA

= ± 16 ∫₂⁻² x dx = ± 16 [x²/2]₂⁻² = ±16

Here we used the fact that the x-component of the field is constant over each face and equal to ±1, depending on the orientation of the face. The other two components of the field do not contribute to the flux through the x-oriented faces.

Since there are six faces, the total flux through the closed surface is Φ = 6(±16) = ±96. By Gauss's Law, this flux is proportional to the charge enclosed by the surface:

Φ = ∫ E · dA = Q/ε0

Solving for Q, we get Q = Φ ε0 = ±96 ε0.

Therefore, the charge enclosed by the cube is ±96 times the permittivity of free space ε0.

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Answer:

I think it's 66,6666666665

Step-by-step explanation:

/5 = is the total of strawberries = 100

/6 = is the total of strawberries (ate) = 80

Arianna :

100 ÷ 5 = 20 × 4 = 80

Arianna's Sister :

5/6 out of 80 [4/5 !] = 80 ÷ 6 = 13,3333333333 × 5 = 66,6666666665

Hope I helped you ! :>

The reason for each step to justify it is:

Division Property of Equality: This step involves dividing both sides of the equation by 4 to isolate the variable term on one side of the equation.

The step involved here is: (w+25)/4

Multiplication Property of Equality: This step involves multiplying both sides of the equation by 4 to simplify the equation and cancel out the denominator.
The step involved here is: 4*(w+25)

Addition Property of Equality: This step involves subtracting 25 from both sides of the equation to isolate the variable term on one side of the equation.

The step involved here is: w+25

Simplifying: This step involves simplifying the expression by combining like terms and performing arithmetic operations to simplify the equation.

The step involved here is: w+25-25=12-25

Subtraction Property of Equality: This step involves subtracting 25 from both sides of the equation to isolate the variable term and solve for the variable.

The step involved here is: w=-13

What is meant by division?

Division is a mathematical operation that involves splitting a quantity into equal parts or groups. It is represented by the symbol ÷ or / and is the inverse operation of multiplication. Division can be used to solve problems related to sharing, fractions, and ratios.

What is meant by multiplication?

Multiplication is a mathematical operation that involves adding a number to itself a certain number of times. It is represented by the symbol × or * and is the inverse operation of division. Multiplication can be used to solve problems related to scaling, area, volume, and rates.

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To simulate a coin tossing environment using a random function, you can follow these steps:

1. Import the random module: `import random`
2. Initialize variables to count the number of heads and tails: `heads = 0` and `tails = 0`
3. Use a for loop to toss the coin 100 times: `for i in range(100):`
4. In the loop, generate a random number between 0 and 1: `toss = random.randint(0, 1)`
5. If the toss is 0, increment the heads counter, otherwise increment the tails counter: `heads += toss == 0` and `tails += toss == 1`
6. After the loop, print the number of heads and tails.

For Part 2, you can use the `randint()` function from the random module to generate random integers within specific ranges:

a) `n = random.randint(1, 3)`
b) `n = random.randint(1, 150)`
c) `n = random.randint(0, 11)`
d) `n = random.randint(1000, 3112)`
e) `n = random.randint(-1, 3)`
f) `n = random.randint(-4, 15)`

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Answer:

8.3[tex]cm^{3}[/tex]

Step-by-step explanation: