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bija089 [108]
3 years ago
9

If f(x) = x-5 and g(x) = 3x2-1, what is (fºg)(x)?

Mathematics
1 answer:
Kobotan [32]3 years ago
4 0

Answer:

<h2>(f\circ \: g)(x) =  {3x}^{2}  - 6</h2>

Step-by-step explanation:

f(x) = x - 5

g(x) = 3x² - 1

To find (fºg)(x) substitute g(x) into f(x) that's for every x in f (x) replace it with g (x)

That's

<h3>(f\circ \: g)(x) = 3 {x}^{2}  -1 - 5</h3>

We have the final answer as

<h3>(f\circ \: g)(x) =  {3x}^{2}  - 6</h3>

Hope this helps you

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If a drug has a concentration of 8.22 mg per 3.039 mL, how many mL are needed to give 7.469 gram of the drug? Round to 1 decimal
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7 0
3 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.

5 0
2 years ago
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