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2 years ago

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After constructing a confidence interval estimate for a population mean, you believe that the interval is useless because it is

too wide. In order to correct this problem, you need to: a. increase the sample size b. increase the sample mean c. increase the confidence coefficient d. decrease the sample size e. increase the population size

1 answer:

choli [55]2 years ago

8 0

Answer:

a. increase the sample size

Step-by-step explanation:

The width of a confidence interval is twice the margin of error, which is given by:

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standrd deviation of the population, n is the size of the sample, and z is the confidence coefficient(the higher the confidence level, the higher z is).

The only of those variables which is direct proportional to M is the sample size. This means that if we want to decrease the width of M, we need to increase the sample size.

So the correct answer is:

a. increase the sample size

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Zac and Lynn are each traveling on a trip. So far, Zac has traveled 123.75 miles in 2.25 hours. Lynn leaves half an hour after Z

Savatey [412]

Complete question is;

Zac and Lynn are each traveling on a trip. So far, Zac has traveled 123.75 miles in 2.25 hours. Lynn leaves half an hour after Zac. So far

she has traveled 105 miles in 1.75 hours. Assume Zac and Lynn travel at constant rates.

Let a represent the number of hours that have elapsed since Zac started traveling. Let y represent the number of miles traveled. Write a system of linear equations that represents the distance each of them has traveled since Zac left on his trip.

Assume Zac and Lynn continue to travel at the same constant rates and make no stops.

Determine the solution of the system of linear equations.

Answer:

Zac: y = 55a

Lynn: y = 60(a - ½)

6 hours after Zac started traveling, both Zac and Lynn would have covered 330 miles each.

Step-by-step explanation:

Zac has traveled 123.75 miles in 2.25 hours. Since he travels at constant speed, we can say;

zac's speed = 123.75/2.25 = 55 mi/hr

Similarly, Lynn traveled 105 miles in 1.75 hours. Thus, since she travels at a constant speed;

Lynn's speed = 105/1.75 = 60 mi/hr

Now, we are told that a represents the number of hours that have elapsed since Zac started traveling and y represents the number of miles traveled.

Thus;

a hours after Zac started travelling, his distance covered will be;

Zac: y = 55a

Now,for Lynn, since she started ½ an hour after Zac, it means a hours after Zac started, she had traveled (a - ½) hours.

Thus, Lynn's distance traveled after Zac started = 60(a - ½)

Lynn: y = 60(a - ½)

The solution will be when they have travelled equal distances a hours after Zac started. Thus;

55a = 60(a - ½)

55a = 60a - 30

60a - 55a = 30

5a = 30

a = 30/5

a = 6 hours

Putting 6 for a in y = 55a, we have;

y = 55 × 6

y = 330 miles

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Answer:

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Step-by-step explanation:

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

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2 years ago

The diameter of Circle Q terminates on the circumference of the circle at (0,3) and (0,-4). Write the equation of the circle in

Gnesinka [82]

First, determine the center of the circle by getting the midpoint of the points given for the circumference.
midpoint = ((0 + 0)/2, (3 + -4)/2)
midpoint (0, -0.5)
Then, we get the radius by determining the distance from either of the circumferential point to the center.
radius = √(0 - 0)² + (3 +4)² = 7
The equation for the circle would be,
x² + (y + 0.5)² = 7²

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2 years ago