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solmaris [256]
3 years ago
15

A ladder 10 ft long rests against a vertical wall. Let theta be the angle between the top of the ladder and the wall and let be

the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to theta when theta=pi/3?
Mathematics
1 answer:
IgorC [24]3 years ago
4 0

Answer:

5 ft/s

Step-by-step explanation:

Let L be the length of the ladder. The relation between the angle and distance X is:

X = L * sin θ      the change of X will be:

dX/dt = L * cos θ    If we evaluate the expression when θ = π/3, we get the change of X:

dX/dt = 10 * cos(π/3)

dX/dt = 5 ft/s

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inysia [295]

Answer:

<u>C) −2x − 2y + 17</u>

Step-by-step explanation (PEMDAS):

First, we do <u>3 (y + 5)</u> because of the parenthesis:

6x − 5y + 2 − 8x + 3y + 15

Since there are no exponents, multiplication ,or division, we will add and subtract. But we have to make sure we <u>combine like terms</u>. X with X's, Y with Y's and constants with constants. And ALWAYS go from left to right

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8 0
4 years ago
A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life
alisha [4.7K]

Answer:

a)

The null hypothesis is H_0: \mu \leq 10

The alternative hypothesis is H_1: \mu > 10

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

H_0: \mu \leq 10

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

H_1: \mu > 10

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that \mu = 10

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that n = 45, X = 10.5, s = 1.8

Then

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}

t = 1.86

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

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