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3 years ago

PLEASE HELP ITS A FUNCTION

Mathematics

1 answer:

Agata [3.3K]3 years ago

5 0

f(3) is equal to 23.

In order to solve for f(3), you must put 3 in for all values of x. This is what the notation of the 3 being in parenthesis means. So we take the original and place 3 in every time we see x.

f(x) = $2x^{2} + 5 \sqrt{x - 2}$

f(3) = $2(3)^{2} + 5 \sqrt{(3) - 2}$

f(3) = $2(9) + 5 \sqrt{1}$

f(3) = $18 + 5(1)$

f(3) = $18 + 5$

f(3) = $23$

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textRequest reports that adults 18–24 years old send and receive 128 texts every day. Suppose we take a sample of 25–34 year old

Butoxors [25]

Testing the hypothesis, we have that:

The null hypothesis is: $H_0: \mu = 128$

The alternative hypothesis is: $H_1: \mu \neq 128$

b) The p-value of the test is of 0.1212.

c) Since the p-value of the test is of 0.1212 > 0.05, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

d) Since |z| = 1.55 < 1.96, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item a:

At the null hypothesis, we test if the mean is the same, that is, of 128 texts every day, hence:

$H_0: \mu = 128$

At the alternative hypothesis, we test if the mean is different, that is, different of 128 texts every day, hence:

$H_1: \mu \neq 128$

Item b:

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the population.

n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 118.6, \mu = 128, \sigma = 33.17, n = 30$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{118.6 - 128}{\frac{33.17}{\sqrt{30}}}$

$z = -1.55$

Since we have a two-tailed test, as we are testing if the mean is different of a value, the p-value is P(|z| < 1.55), which is 2 multiplied by the p-value of z = -1.55.

Looking at the z-table, z = -1.55 has a p-value of 0.0606

2(0.0606) = 0.1212

The p-value of the test is of 0.1212.

Item c:

Since the p-value of the test is of 0.1212 > 0.05, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item d:

Using a z-distribution calculator, the critical value for a two-tailed test with 95% confidence level is |z| = 1.96.

Since |z| = 1.55 < 1.96, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

A similar problem is given at brainly.com/question/25369247

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3 years ago

What is 40/100 simplified

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Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100
Common Factors: 1, 2, 4, 5, 10, 20
Greatest Common Factor: 20

$\frac{40:20}{100:20}= \frac{2}{5} \\ \\ \frac{40}{100}= \frac{2}{5}$

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4 years ago

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Answer:

Step-by-step explanation:

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3 years ago

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