Factor the expression completely. 3x3

The scale on a map says the 6 cm represents 15 km. What is the actual number of kilometers that is represented by 11 centimeters

Sergio039 [100]

Step-by-step explanation:

15 ÷6= 2.5

2.5×11=275km

4 0

3 years ago

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:

a)

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

4 0

3 years ago

What is 6 1/3−1 2/3?

kolbaska11 [484]

Answer:

4 2/3!

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Below is the graph of a trigonometric function. It has a minimum point at (2, -0.9) and an amplitude of 3.