find the area of the region that is bounded by the curve r=5sin(θ)−−−−−−√ and lies in the sector 0≤θ≤π

The P-value for a one-tailed z-test with Ha: p > 0.4 and z = 1.49 is 0.0675, indicating insufficient evidence to reject the null hypothesis at the 0.05 level of significance.

To find the P-value for a z-test with Ha: p > 0.4 and z = 1.49, we first calculate the corresponding area under the standard normal distribution curve.

Using a standard normal table or a calculator, we find that the area to the right of z = 1.49 is 0.0675.

Since the alternative hypothesis is one-tailed, the P-value is equal to the area in the tail to the right of z = 1.49.

Therefore, the P-value for this test is 0.0675 or 6.75% (rounded to four decimal places).

This means that if the null hypothesis is true, there is a 6.75% chance of observing a sample proportion as extreme as or more extreme than the one we obtained.

Since the P-value (6.75%) is greater than the significance level (α), we fail to reject the null hypothesis at the α = 0.05 level of significance. We do not have sufficient

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Equation of the tangent plane is: 12(x-6) + 4(y-2) - 14(z-7) = 0

Correct answer is option C.

We need to first find the partial derivatives of the given surface with respect to x, y, and z.

∂f/∂x = 2x

∂f/∂y = 2y

∂f/∂z = -2z

Then, we can evaluate them at the given point (6, 2, 7):

∂f/∂x = 2(6) = 12

∂f/∂y = 2(2) = 4

∂f/∂z = -2(7) = -14

Equation of the tangent plane is;

12(x-6) + 4(y-2) - 14(z-7) + D = 0

where D is the constant we need to find by plugging in the point (6, 2, 7):

12(6-6) + 4(2-2) - 14(7-7) + D = 0

D = 0

Equation of the tangent plane is:

12(x-6) + 4(y-2) - 14(z-7) = 0

So the correct answer is option C.

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y = sin(t) is indeed a solution to the differential equation (dy/dt)² = 1 - y².

We'll need to perform the following steps:

Step 1: Find the derivative of y with respect to t.
Step 2: Square the derivative and substitute it into the equation.
Step 3: Check if the equation holds true with the given function.

Step 1:
To find the derivative of y = sin(t) with respect to t, we use the basic differentiation rule for sine:
(dy/dt) = cos(t).

Step 2:
Next, we square the derivative:
(dy/dt)² = cos²(t).

Step 3:
Now we substitute this expression and y = sin(t) into the given equation:
(cos²(t)) = 1 - (sin²(t)).

Using the trigonometric identity sin²(t) + cos²(t) = 1, we can see that the equation holds true:
cos²(t) = 1 - sin²(t).

Thus, y = sin(t) is indeed a solution to the differential equation (dy/dt)² = 1 - y².

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The approximate P-values for the test statistics are: a) 0.045, b) 0.104, and c) 0.996.

To calculate the P-value for each test statistic, we use the chi-square distribution with degrees of freedom (df) equal to n-1.

a) For x2_0 = 25.2 and n = 20, df = 19. Using a chi-square table or calculator, we find the P-value is approximately 0.045.
b) For x2_0 = 15.2 and n = 12, df = 11. The P-value is approximately 0.104.
c) For x2_0 = 4.2 and n = 15, df = 14. The P-value is approximately 0.996.

The P-values help us determine whether to reject or fail to reject the null hypothesis (H0: σ2 = 5) in favor of the alternative hypothesis (H1: σ2 < 5). The smaller the P-value, the stronger the evidence against H0.

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We have computed the Taylor polynomials of the given function f (x) = cos (4x), using around 6 decimals for approximation. These polynomials can then be used to approximate the given function.

Function is a block of code that performs a specific task. It can accept input parameters and return a value or a set of values. Functions are used to break down a complex problem into simple, manageable tasks. They also help improve code readability and re-usability. By using functions, you can write code more efficiently and easily maintain your program.

The Taylor series of a given function is a polynomial approximation of that function, derived using derivatives. In this case, we are asked to compute the Taylor polynomial for the function f (x) = cos (4x).

The Taylor polynomials of f are as follows:

p0(x) = 1

p1(x) = 1 - 8x2

p2(x) = 1 - 8x2 + 32x4

p3(x) = 1 - 8x2 + 32x4 - 128x6

p4(x) = 1 - 8x2 + 32x4 - 128x6 + 512x8

For any approximations, we can use around 6 decimals. For instance, if x = 0.5, then p4(0.5) = 0.988377, which is an approximation of the actual value of f (0.5), which is 0.98879958.

In conclusion, we have computed the Taylor polynomials of the given function f (x) = cos (4x), using around 6 decimals for approximation. These polynomials can then be used to approximate the given function.

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The probability that a randomly selected person will have an IQ score of less than 80 is approximately 0.0918, or 9.18%

To find the probability that a randomly selected person will have an IQ score of less than 80, we need to consider the properties of the normal distribution, as IQ scores typically follow a normal distribution with a mean (μ) of 100 and a standard deviation (σ) of 15.

1. Calculate the z-score: The z-score represents the number of standard deviations a data point is from the mean. Use the formula:

z = (X - μ) / σ

where X is the IQ score, μ is the mean, and σ is the standard deviation.

z = (80 - 100) / 15
z = -20 / 15
z = -1.3333

2. Look up the z-score in a standard normal distribution table or use a calculator to find the corresponding probability. In this case, the probability is 0.0918.

Therefore, the probability that a randomly selected person will have an IQ score of less than 80 is approximately 0.0918, or 9.18% when rounded to four decimal places.

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For a one-sample t-test with df=14, there were 15 participants in the study. The degrees of freedom in any one group are equal to the sample size minus 1.


a. The hypotheses for the study are:
- Null hypothesis (H0): The mean number of racist remarks in the school is equal to the population mean (µ=7).
- Alternative hypothesis (H1): The mean number of racist remarks in the school is different from the population mean (µ≠7).

b. Using a t-table or calculator, with df=9 (10 classes - 1) and alpha=.05 (two-tailed), the critical t-value is approximately ±2.262.

c. To compute the one-sample t-test:
1. Find the sample mean (M) and sample standard deviation (s) of the observed racist remarks.
2. Compute the t-value using the formula: t = (M - µ) / (s / √n), where n is the sample size.

d. In APA format, report the t-value, degrees of freedom, and the direction of the results, for example:

"A one-sample t-test revealed a significant difference in the number of racist remarks at the school compared to the population mean, t(9) = X.XX, p < .05 (or p = Y.YY, if you have the exact p-value), with fewer racist remarks observed." (Replace X.XX and Y.YY with the calculated values.)

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Answer:

C. The zeros of f(x) are -1 and 1.

When each point on a plane of a two-dimensional coordinate system is decided by a distance from a reference point and an angle is taken from a reference direction, it is known as the polar coordinate system.

Polar coordinates are advantageous when the region being integrated over has a circular or symmetric shape. This is because polar coordinates use angles and radii to describe points in a two-dimensional plane, which aligns well with circular and symmetric shapes. Additionally, polar coordinates can simplify the integrand, as some functions are more easily expressed in terms of angles and radii rather than Cartesian coordinates.

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The given equation r = 2 sin θ represents a curve in polar coordinates. To identify the surface whose equation is given, we need to convert the equation into rectangular coordinates.

To convert the equation, we use the relationships between polar and rectangular coordinates:

x = r cos θ
y = r sin θ

Substituting the given value of r = 2 sin θ, we get:

x = 2 sin θ cos θ
y = 2 sin² θ

Simplifying these equations, we get:

x = sin 2θ
y = 2 sin² θ

The resulting equations represent a surface known as a "lemniscate of Bernoulli." It is a closed, symmetric curve with two loops, resembling the shape of a figure-eight. The lemniscate of Bernoulli is named after the Swiss mathematician Jacob Bernoulli, who first studied the curve in the 17th century.

In summary, the surface whose equation is given by r = 2 sin θ is a lemniscate of Bernoulli, which can be represented by the equations x = sin 2θ and y = 2 sin² θ in rectangular coordinates.

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For the given equation there are 3 real solutions they are -4/3, -3, 1 , under the condition that the given equation is  x² = 1/(x+3)

The equation x²= 1/(x+3) can be restructured as
x³ + 3x² - 1 = 0.
This is a cubic equation and could be evaluated applying the cubic formula. Then, we can also apply the rational root theorem to search the rational roots of the equation.
The rational root theorem projects that if a polynomial equation has integer coefficients, then any rational root of the equation should be of the form p/q
Here,
p = factor of the constant term and q is a factor of the leading coefficient.
For the given case,
the constant term is -1 and the leading coefficient is 1.
Hence, any rational root of the equation should be of the form p/q
Here, p is a factor of -1 and q is a factor of 1.
The possible rational roots are ±1 and ±1/3.
Applying the principle of testing these values, we evaluate that
x = -1/3 is a root of the equation.
Then, we can factorize
x³ + 3x² - 1 as (x + 1/3)(x² + 2x - 3).
The quadratic factor can be simplified further as
(x + 3)(x - 1),
Then, the solutions to the original equation are
x = -4/3, x = -3, and x = 1.
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When dealing with exponential functions given by y = (a + c)^x, where the constant 'c' is used to achieve horizontal shifts, there are particular effects on the domain, range, and asymptotes

The function's output values, or range, persist unchanged since it can assume any positive value for input from the vertical axis. Similarly, factorizing by adding constants does not impact the function's input values, otherwise known as the domain.

While horizontally shifting the exponentially-decreasing function, its horizontal asymptote remains unaffected; however, the positional shift depends on the magnitude and direction of said diasporic events. Equivalently, rightward shifts append positively and leftward motions take away from the aforementioned translation distance.

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The value at t0 of a 4-year annuity depends on the payment amount and the interest rate. Assuming the interest rate is 5%, the value of each of the following 4-year annuities can be calculated using the present value of an annuity formula.

In summary, at t0, the value of each 4-year annuity is approximately $36,376 for an annuity that pays $10,000 at the end of each year, $36,252 for an annuity that pays $5,000 at the end of each half-year, $36,172 for an annuity that pays $1,000 at the end of each quarter, and $36,130 for an annuity that pays $500 at the end of each month, assuming a 5% interest rate. For each annuity, the present value of an annuity formula was used to compute the value at t0, and the interest rate was changed based on the frequency of payments.

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The diameter of the columns follows a normal distribution with a mean (μ) of 8.25 inches and a standard deviation (σ) of 0.1 inch.

For the distribution of X, the mean will be the same as the population mean, which is 8.25 inches, and the standard deviation will be the population standard deviation divided by the square root of the sample size (n):
Standard deviation of X = σ/√n = 0.1/√35 ≈ 0.0169 inches

So, the distribution of the sample mean diameter (X) is approximately N(8.25, 0.0169²).
Z = (X - μ) / (σ/√n) = (8 - 8.25) / 0.0169 ≈ -14.7929
However, since this Z-score is extremely large in magnitude, the probability is very close to 1 (almost certain) that the sample mean diameter for the 35 columns will be greater than 8 inches.

Using a standard normal table or calculator, we can find that the probability of getting a z-score of -14.88 or lower is practically 0. Therefore, the probability that the sample mean diameter for the 35 columns will be greater than 8 inches is practically 1.

The diameter of the columns follows a normal distribution with a mean (μ) of 8.25 inches and a standard deviation (σ) of 0.1 inch. When a sample of 35 columns is taken, we can find the distribution of the Sample Size diameter (X) using the Central Limit Theorem.

For the distribution of X, the mean will be the same as the population mean, which is 8.25 inches, and the standard deviation will be the population standard deviation divided by the square root of the sample size (n):

Standard deviation of X = σ/√n = 0.1/√35 ≈ 0.0169 inches

So, the distribution of the sample mean diameter (X) is approximately N(8.25, 0.0169²).

Z = (X - μ) / (σ/√n) = (8 - 8.25) / 0.0169 ≈ -14.7929

However, since this Z-score is extremely large in magnitude, the probability is very close to 1 (almost certain) that the sample mean diameter for the 35 columns will be greater than 8 inches.

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The value of Fc[f(x)] is: (1 - z²)/(1 + z²)²

The given function is f(x) = xeˣ.

The Fourier Sine Transform of f(x) is given by:

Fs[f(x)] = ∫₀^∞ f(x) sin(zx) dx

Taking the derivative of f(x) with respect to x, we get:

f'(x) = (x + 1) eˣ

Taking the Fourier Sine Transform of f'(x), we get:

Fs[f'(x)] = ∫₀^∞ f'(x) sin(zx) dx

= ∫₀^∞ (x + 1) eˣ sin(zx) dx

Using integration by parts, we get:

Fs[f'(x)] = [(x + 1) (-cos(zx))/z - eˣ sin(zx)/z]₀^∞

+ (1/z) ∫₀^∞ eˣ cos(zx) dx

Simplifying the above expression, we get:

Fs[f'(x)] = 2z/(z² + 1)²

The Fourier Cosine Transform of f(x) is given by:

Fc[f(x)] = ∫₀^∞ f(x) cos(zx) dx

Using integration by parts, we get:

Fc[f(x)] = [xeˣ sin(zx)/z + eˣ cos(zx)/z²]₀^∞

- (1/z²) ∫₀^∞ eˣ sin(zx) dx

Since eˣ sin(zx) is an odd function, the integral on the right-hand side is the Fourier Sine Transform of eˣ sin(zx), which we have already calculated as 2z/(z² + 1)². Substituting this value in the above expression, we get:

Fc[f(x)] = (1 - z²)/(1 + z²)²

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the quadrilateral with vertices A(0, 0), B(4, 3), C(13, -9), and D(9, -12) is a parallelogram and a rhombus

what is  quadrilateral ?

A quadrilateral is a polygon with four sides and four vertices. It is a type of geometric shape that can have various properties and characteristics depending on the lengths of its sides, the angles between those sides, and the positions of its vertices.

In the given question,

To graph the quadrilateral, we can plot the given points on a coordinate plane and connect them in order.

Quadrilateral ABCD

To prove what type of quadrilateral it is, we can use both slope and distance measurements.

First, we can calculate the slopes of each side of the quadrilateral:

Slope of AB: (3 - 0)/(4 - 0) = 3/4

Slope of BC: (-9 - 3)/(13 - 4) = -12/9 = -4/3

Slope of CD: (-12 - (-9))/(9 - 13) = -3/-4 = 3/4

Slope of DA: (0 - (-12))/(0 - 9) = 12/9 = 4/3

We can see that the slopes of opposite sides are equal: AB and CD have the same slope of 3/4, and BC and DA have the same slope of -4/3. This tells us that the quadrilateral is a parallelogram.

Next, we can calculate the distances of each side of the quadrilateral:

Distance between A and B: √((4 - 0)² + (3 - 0)²) = √(16 + 9) = √25 = 5

Distance between B and C: √((13 - 4)² + (-9 - 3)²) = √(81 + 144) = √225 = 15

Distance between C and D: √((9 - 13)² + (-12 - (-9))²) = √(16 + 9) = √25 = 5

Distance between D and A: √((0 - 9)² + (0 - (-12))²) = √(81 + 144) = √225 = 15

We can see that opposite sides have the same length: AB and CD have a length of 5, and BC and DA have a length of 15. This tells us that the parallelogram is also a rhombus.

Therefore, we have proved that the quadrilateral with vertices A(0, 0), B(4, 3), C(13, -9), and D(9, -12) is a parallelogram and a rhombus

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From the histogram, we can say that the observations were selected from a normal distribution. We are 95% confident that the population mean ACT score for students taking freshman calculus is between 24.208 and 26.582. The calculus students have a higher average score of 25.395. we are 99% confident that the population standard deviation is between 8.246 and 23.639.

To check whether the observations were selected from a normal distribution, we can create a histogram or a normal probability plot.

From the histogram, it seems plausible that the observations were selected from a normal distribution, as the data appears to be roughly symmetric.

Using the given data, we can calculate a 95% confidence interval for the population mean using the formula

confidence interval = sample mean ± (critical value)(standard error)

The critical value for a 95% confidence interval with 19 degrees of freedom (n - 1) is 2.093.

The sample mean is 25.395, and the standard error can be calculated as the sample standard deviation divided by the square root of the sample size

standard error = 2.630 / sqrt(20) = 0.588

Therefore, the 95% confidence interval is

25.395 ± (2.093)(0.588)

= [24.208, 26.582]

We are 95% confident that students taking freshman calculus is between 24.208 and 26.582.

The university ACT average for entire freshmen that year was about 21. The calculus students have a higher average score of 25.395. Therefore, we can say that the calculus students performed better on the ACT than the average freshman.

To calculate a 99% confidence interval for the population standard deviation, we can use the chi-square distribution. The formula for the confidence interval is

confidence interval = [(n - 1)s^2 / χ^2_(α/2), (n - 1)s^2 / χ^2_(1-α/2)]

where n is the sample size, s is the sample standard deviation, and χ^2_(α/2) and χ^2_(1-α/2) are the chi-square values with α/2 and 1-α/2 degrees of freedom, respectively.

For a 99% confidence interval with 19 degrees of freedom, the chi-square values are 8.906 and 32.852.

Plugging in the values from the sample, we get

confidence interval = [(19)(6.615^2) / 32.852, (19)(6.615^2) / 8.906]

= [8.246, 23.639]

Therefore, we are 99% confident  that the population standard deviation is between 8.246 and 23.639. This interval assumes that the population is normally distributed. If the population is not normally distributed, the interval may not be valid.

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New pump:

[tex]\text{5 hours = 1 pool}[/tex]

[tex]\text{1 hour} = \dfrac{1}{5} \ \text{pool}[/tex]

Old pump:

[tex]\text{15h = 1 pool}[/tex]

[tex]\text{1h} = \dfrac{1}{15} \ \text{pool}[/tex]

If both work together

[tex]\text{1h} = \dfrac{1}{5}+ \dfrac{1}{15}= \dfrac{4}{15} \ \text{pool}[/tex]

[tex]\dfrac{4}{15} \ \text{pool = 1 hour}[/tex]

[tex]\dfrac{1}{15} \ \text{pool} = \dfrac{1}{4} \ \text{hour}[/tex]

[tex]\dfrac{15}{15} \ \text{pool} =\dfrac{1}{4} \times15[/tex]

1 pool = 3.75 hours or 3 hours 45 mins

Answer:

38% of 94 is 38.72 simplified

Answer:

38% of 94 is 38.72 simplified

B) 30 - 75x

This is because of the distributive property.

In the expression 3(10-25x) you have to multiply 3 by the numbers inside.

3 x 10 = 30

3 x 25x = 75x

Then, we keep the subtraction sign. The final answer is 30 - 75x.

For the equation [tex]f(x)=x^(12)h(x)[/tex], the value of f(−1) is 7.

To find f(-1), we need to evaluate the function f(x) at x = -1. We are given that [tex]f(x) = x^12 * h(x)[/tex], and [tex]h(-1) = 4[/tex]. Therefore, we can compute f(-1) as follows:

[tex]f(-1) = (-1)^12 * h(-1)[/tex]

[tex]= 1 * h(-1)[/tex]

[tex]= 4[/tex]

We are also given that h(-1) = 7, so we can substitute this value to obtain:

[tex]f(-1) = 1 * h(-1)[/tex]

[tex]= 1 * 7[/tex]

[tex]= 7[/tex]

Therefore, [tex]f(-1) = 7.[/tex]

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In this case, we are defining a linear transformation T from a subspace of dimension 3 (Ra) to a subspace of dimension b (Rb) using the matrix A.

Since A is a 5 x 3 matrix, it maps a vector in Ra (which has dimension 3) to a vector in R5 (which has dimension 5). To determine the dimensions of Ra and Rb, we need to look at the dimensions of the vector x and the matrix A. Since A has 3 columns, the vector x must have 3 entries, so Ra is a subspace of R3. Since T(x) is a vector in R5, b must be 5.

Therefore, we have a = 3 and b = 5. The linear transformation T maps vectors in Ra to vectors in R5, and is defined by T(x) = Ax where A is a 5 x 3 matrix.

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Answer:

Step-by-step explanation:

  A = [tex]\left[\begin{array}{cc}4&-4\\3&-2\end{array}\right][/tex]

3B = [tex]\left[\begin{array}{cc}12&12\\0&3\end{array}\right][/tex]

4 + 12 = 16 ; 12 + ( - 4) = 8

3 + 0 = 3  ; - 2 + 3 = 1

A + 3B = [tex]\left[\begin{array}{cc}16&8\\3&1\end{array}\right][/tex]

[tex](A+3B)^{-1}[/tex] = [tex]\left[\begin{array}{cc}-\frac{1}{8} &1\\\frac{3}{8} &-2\end{array}\right][/tex]

X = C ÷ ( A + 3B ) = C × [tex](A+3B)^{-1}[/tex]

X = [tex]\left[\begin{array}{cc}-1&0\\5&2\end{array}\right][/tex] × [tex]\left[\begin{array}{cc}-\frac{1}{8} &1\\\frac{3}{8} &-2\end{array}\right][/tex] = [tex]\left[\begin{array}{cc}\frac{1}{8} &-1\\\frac{1}{8} &1\end{array}\right][/tex]  

After calculating the area of the cross section which is the sum of the areas of the square and trapezoid,the result is Rounded to the nearest whole number, the area is 120 ft², which corresponds to option D

A trapezoid is a quadrilateral with one pair of parallel sides. The other pair of sides may or may not be parallel.

A cross section is the shape or profile that is obtained when a solid object is cut perpendicular to a particular axis or plane, revealing its internal structure or composition.

According to the given information :

The cross-section formed by slicing a square pyramid parallel to its base consists of a smaller square and a trapezoid. The dimensions of the smaller square are given as 4 feet by 4 feet. To find the area of the trapezoid, we need to calculate the lengths of its two parallel sides. These are the diagonals of the square pyramid base and are equal to √(10² + 10²) = 10√2 feet.

The distance between the two parallel sides is the height of the frustum (portion of the pyramid left after slicing), which is 12 - 4 = 8 feet. Using the formula for the area of a trapezoid, A = 1/2 (b1 + b2)h, where b1 and b2 are the lengths of the parallel sides and h is the distance between them, we get:

A = 1/2 (10√2 + 10√2) x 8 = 80√2

Therefore, the total area of the cross-section is the sum of the areas of the square and trapezoid:

A = 4² + 80√2 ≈ 120.4 ft²

Rounded to the nearest whole number, the area is 120 ft², which corresponds to option D

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The basis for the subspace W = Span(S) is 2.

To use the Minimizing Theorem (Basis for a Subspace Version) to find a basis for the subspace W = Span(S), we first need to create an augmented matrix with the vectors in S and row reduce it to its reduced row echelon form (RREF).

The augmented matrix is:

[5 -3 6 7 | 0]
[3 -1 4 5 | 0]
[7 -5 8 9 | 0]
[1 3 -1 1 | 0]
[1 3 -9 -7 | 0]

Row reducing this matrix to its RREF, we get:

[1 0 1 1 | 0]
[0 1 -2 -1 | 0]
[0 0 0 0 | 0]
[0 0 0 0 | 0]
[0 0 0 0 | 0]

From the RREF, we see that the first two columns correspond to the pivot columns, and the other two columns correspond to the free columns. So, a basis for W is given by the vectors in S that correspond to the pivot columns, which are:

{(5,-3,6,7), (3,-1,4,5)}

Therefore, a basis for W is {(5,-3,6,7), (3,-1,4,5)}, and dim(W) = 2.

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The average cost per month of owning an iPhone decreases as the number of months of ownership increases. After 8 months, the average cost per month is $145.

Assuming a constant monthly service fee of $70, the total cost (C) of owning an iPhone for n months can be calculated as:

C = 600 + 70n

where n is the number of months of ownership.

Using this formula, we can calculate the total cost of owning an iPhone after:

i. 2 months:

C = 600 + 70(2) = 740

ii. 4 months:

C = 600 + 70(4) = 880

iii. 6 months:

C = 600 + 70(6) = 1020

iv. 8 months:

C = 600 + 70(8) = 1160

To find the average cost per month, we can divide the total cost by the number of months:

i. Average cost per month after 2 months: 740 / 2 = 370

ii. Average cost per month after 4 months: 880 / 4 = 220

iii. Average cost per month after 6 months: 1020 / 6 = 170

iv. Average cost per month after 8 months: 1160 / 8 = 145

Therefore, the average cost per month of owning an iPhone decreases as the number of months of ownership increases. After 8 months, the average cost per month is $145.

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The correct answer is option c. i.e. B = 37.2°, b = 264.3, c = 326.9.

To solve the triangle, we can use the given information:
1. A = 11.2°
2. C = 131.6°
3. a = 84.9

Step 1: Find angle B.
Since the sum of angles in a triangle is 180°, we can calculate angle B as follows:
B = 180° - (A + C) = 180° - (11.2° + 131.6°) = 180° - 142.8° = 37.2°

Step 2: Find side b.
We can use the Law of Sines to find side b.
a / sin(A) = b / sin(B)
84.9 / sin(11.2°) = b / sin(37.2°)

Now, solve for b:
b = (84.9 * sin(37.2°)) / sin(11.2°) ≈ 264.3

Step 3: Find side c.
Again, we can use the Law of Sines to find side c.
a / sin(A) = c / sin(C)
84.9 / sin(11.2°) = c / sin(131.6°)

Now, solve for c:
c = (84.9 * sin(131.6°)) / sin(11.2°) ≈ 326.9

So, the final answer is:
B = 37.2°, b = 264.3, c = 326.9, which corresponds to option c.

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To graph the quadratic function y=-2x^2 using the given values of x, one can create a table with two columns: one for x and the other for y. Starting with x=-2, we can substitute this value into the equation to find the corresponding value of y, which is y=-8. Similarly, by substituting -1, 0, 1, and 2 into the equation, we can find corresponding values of y as 2, 0, -2, and -8, respectively. By plotting these points on a graph and connecting them, we get a downward facing parabola with its vertex at (0,0).

Chrissy can knit 0.5 scarves in one day. It would take her 6 days to knit 3 scarves working alone.

Here, Using the unitary method

Lauren can knit 3 scarves in 2 days, which means her rate is 3/2 scarves per day. When Lauren and Chrissy work together, they can knit the same 3 scarves in 1.5 days, which means their combined rate is 2 scarves per day.

We want to find out how long it would take Chrissy to knit 3 scarves working alone.

Let the number of days it takes Chrissy to knit 3 scarves be d. Then her rate is 3/d scarves per day. We know that when Chrissy and Lauren work together, their combined rate is 2 scarves per day, so we can set up the equation

3/2 + 3/d = 2

Multiplying both sides by 2d, we get

3d + 6 = 4d

Simplifying, we get

d = 6

Therefore, it would take Chrissy 6 days to knit 3 scarves working alone.

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