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garik1379 [7]
3 years ago
6

I. You withdraw $17 from your account. II. A running back loses 17 yards in a game. III. The temperature rises 17 degrees. The i

nteger -17 would BEST represent which of these events? A) I only B) III only C) I and II only D) II and III only
Mathematics
1 answer:
Anika [276]3 years ago
4 0

Withdrawing money from an account is a negative value, losing yards in a game would also be a negative value.

A rise in temperature would be a positive value.

Both 1 and 2 could be written as -17.

The answer is C.

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DanielleElmas [232]

Option A:

(x-5)^{2}+(y-3)^{2}=16

Solution:

Given data:

Center of the circle is (5, 3).

Radius of the circle = 4

To find the equation of the circle:

The general form of the equation of a circle in centre-radius format is

(x-h)^{2}+(y-k)^{2}=r^{2}

where (h, k) is the centre of the circle and r is the radius of the circle.

Substitute the given values in the equation of a circle formula:

(x-5)^{2}+(y-3)^{2}=4^{2}

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The equation of the given circle is (x-5)^{2}+(y-3)^{2}=16.

Hence Option A is the correct answer.

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3 years ago
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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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