The age of furaha is 1/2 of the age of her aunt if the sum of their ages is 54 years. find the age of her aunt

Solve p = -44 ÷ 11.<br> GIVING BRAINLIST!!!

IRINA_888 [86]

Answer:

p=-4

Step-by-step explanation:

Can you please answer My math question for 50 points and brainliest?

7 0

2 years ago

Read 2 more answers

How do you do proofs

AnnyKZ [126]

Answer:

you can watch YT videos or search online. search: How to solve proofs. Also search up proof theorems. This will help you even more towards solving more complex proof problems.

Step-by-step explanation:

8 0

3 years ago

A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 100 items, th

pochemuha

Answer:

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

Step-by-step explanation:

A manufacturer considers his production process to be out of control when defects exceed 3%.

At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So

$H_0: p \leq 0.03$

At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So

$H_1: p > 0.03$

The test statistic is:

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.03 is tested at the null hypothesis

This means that $\mu = 0.03, \sigma = \sqrt{0.03*0.97}$

In a random sample of 100 items, the defect rate is 4%.

This means that $n = 100, X = 0.04$

Value of the test statistic:

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

$z = \frac{0.04 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{100}}}$

$z = 0.59$

P-value of the test

The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.

Looking at the z-table, z = 0.59 has a p-value of 0.7224

1 - 0.7224 = 0.2776

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

4 0

2 years ago

In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

1) Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: