A rectangle measures 11 cm by 30 cm what id its area

Need help! Do anyone know?

Katyanochek1 [597]

Could you please show the question so I can try to answer it

4 0

2 years ago

The base of a triangle is 13cm and its height is 5cm. what is the area of the triangle

mario62 [17]

Answer:

32.5cm

Step-by-step explanation:

Using the formula A=h_b*b/2

We get 5*13/2=32.5

3 0

3 years ago

A researcher finds that of 1000 people who said that they attend a religious service at least once a week, A stopped to help a p

Ne4ueva [31]

Answer:

There is enough evidence to support the claim that the proportions are not equal. (P-value: 0.048).

Step-by-step explanation:

The question is incomplete:

<em>"A researcher finds that of 1000 people who said that they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are different."</em>

This is a hypothesis test for the difference between proportions.

The claim is that the proportions are not equal.

Then, the null and alternative hypothesis are:

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

The significance level is 0.05.

The sample 1, of size n1=1000 has a proportion of p1=0.031.

$p_1=X_1/n_1=31/1000=0.031$

The sample 2, of size n2=1200 has a proportion of p2=0.018.

$p_2=X_2/n_2=22/1200=0.018$

The difference between proportions is (p1-p2)=0.013.

$p_d=p_1-p_2=0.031-0.018=0.013$

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

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{31+22}{1000+1200}=\dfrac{53}{2200}=0.024$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.024*0.976}{1000}+\dfrac{0.024*0.976}{1200}}\\\\\\s_{p1-p2}=\sqrt{0+0}=\sqrt{0}=0.007$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.013-0}{0.007}=\dfrac{0.013}{0.007}=1.98$

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

$P-value=2\cdot P(z>1.98)=0.048$

As the P-value (0.048) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportions are not equal.

6 0

3 years ago

Im stuck again I SUCK AT MATH please help!!

hodyreva [135]

Answer:

2 does not belong

Step-by-step explanation:

We know that 2 cups of orange juice make 3 cups of fruit punch. So 2/3 cups of orange juice makes 1 cup of fruit punch. Due to this, you know that 2/3 belongs.

7 0

3 years ago

Suppose we want to see if American children have higher levels of cholesterol than the average child (i.e., in the entire world

lina2011 [118]

Answer:

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$

$p_v =P(t_{(24)}>5.5)=0.00000589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation

$\bar X=201$ represent the mean

$s=10$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =190$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:

Null hypothesis: $\mu \leq 190$

Alternative hypothesis: $\mu > 190$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

Since is a one side test the p value would be:

$p_v =P(t_{(24)}>5.5)=0.00000589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

3 0

3 years ago