Helppp plz. 13+5 times a number X is under 30

Anit [1.1K]

Yes they are equivalent expressions. If you take the z to the power of two, and subtract that from the other z to the power of two, it is zero. So you are left with four which is the answer

6 0

3 years ago

Decimals between 10 and 15 with an interval of 0.75

IgorC [24]

10.75
11.50
11.80
12.55
13.30
14.05
14.80
15.55

3 0

3 years ago

Solve this question peasents nah jk lol

Zina [86]

v = √16 · √d

d = 300 ft

v = √16 · √300

v = √16 · √100 · √3

v = 4 · 10 · √3

<em>v = 40√3 ft/sec </em> <== simplified radical form

That's about 69.3 feet per sec, or 47.2 miles per hour.

5 0

3 years ago

A simple random sample of size has mean and standard deviation.Construct a confidence interval for the population mean.The param

scZoUnD [109]

ANSWER:

EXPLANATION:

A simple random sample of size has mean and standard deviation. Construct a confidence interval for the population mean. The parameter is the population The correct method to find the confidence interval is the method.

The sample size is not given. Mean and Standard Deviation are not given.

To construct a confidence interval for the population mean, first find out the margin of error of the sample mean. This is why you need a confidence interval. If you are 90% confident that the population mean lies somewhere around the sample mean then you construct a 90% confidence interval.

This is equivalent to an alpha level of 0.10

If you are 95% sure that the population mean lies somewhere around the sample mean, your alpha level will be 0.05

In summary, get the values for sample size (n), sample mean, and sample standard deviation.

Make use of a degrees of freedom of (n-1).

7 0

3 years ago

AT&T would like to test the hypothesis that the proportion of 18- to 34-year-old Americans that own a cell phone is less tha

Vera_Pavlovna [14]

Answer:

The null and alternative hypothesis can be written as:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of 18- to 34-year-old Americans that own a cell phone is less than the proportion of 35- to 49-year-old Americans.

This claim will be reflected in the alternnative hypothesis, that will state that the population proportion 1 (18 to 34) is significantly smaller than the population proportion 2 (35 to 49).

On the contrary, the null hypothesis will state that the population proportion 1 is ot significantly smaller than the population proportion 2.

Then, the null and alternative hypothesis can be written as:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

The significance level is assumed to be 0.05.

The sample 1, of size n1=200 has a proportion of p1=0.63.

$p_1=X_1/n_1=126/200=0.63$

The sample 2, of size n2=175 has a proportion of p2=0.68.

$p_2=X_2/n_2=119/175=0.68$

The difference between proportions is (p1-p2)=-0.05.

$p_d=p_1-p_2=0.63-0.68=-0.05$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{126+119}{200+175}=\dfrac{245}{375}=0.653$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.653*0.347}{200}+\dfrac{0.653*0.347}{175}}\\\\\\s_{p1-p2}=\sqrt{0.001132+0.001294}=\sqrt{0.002427}=0.049$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.05-0}{0.049}=\dfrac{-0.05}{0.049}=-1.01$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=P(z$

As the P-value (0.1554) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of 18- to 34-year-old Americans that own a cell phone is less than the proportion of 35- to 49-year-old Americans.

5 0

3 years ago