The height of the rectangle and the triangle are the same. The base of the triangle is half the base of the rectangle.

PLEASE HELP ME! FAST!

zepelin [54]

Average rate is given by,

$\langle r \rangle= \frac{f(b)-f(a)}{b-a}$

for section A, a=1, b=3

hope this helps.

3 0

3 years ago

Help quick i will mark u brainlyest

Mumz [18]

4+48=52
if you divide 52 by 4 you should get 13.
this means michael paid 13 dollars per pizza :)

6 0

3 years ago

The lengths of the sides of a triangle are given below. Classify the triangle as acute, obtuse, or right.

enot [183]

Answer:

B. Obtuse Triangle

Step-by-step explanation:

Use Pythagorean to identify if the triangle is right:

9² = 7² + 3²
81 = 58

We see the triangle is not right and the difference of the squares tells us the angle opposite to side with measure of 9 is greater than 90°.

It means the triangle is obtuse.

Correct choice is B

8 0

3 years ago

Things did no go quite as planned. You invested $19,660 part of it in a stock that paid 12 percent annual interest. However the

Doss [256]

0.05x+0.12 (19660-x)=2060
Solve for x
X=4274.29 at 5%

19,660−4,274.29
=15,385.71 at 12%

8 0

3 years ago

To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin

Murrr4er [49]

Answer:

a) $t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733$

$p_v =2*P(t_{15}>1.733)=0.104$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.

b) $t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6$

$p_v =2*P(t_{15}>2.6)=0.0201$

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, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

$\bar X=31.3$ represent the sample mean

$s=3$ represent the sample standard deviation

$n=16$ sample size

$\mu_o =30$ 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 different from 30, the system of hypothesis are :

Null hypothesis: $\mu = 30$

Alternative hypothesis: $\mu \neq 30$

Since we don't know the population deviation, is better apply a z 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:

$t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733$

P-value

First w eneed to find the degrees of freedom given by:

$df=n-1=16-1 =15$

Since is a two-sided test the p value would given by:

$p_v =2*P(t_{15}>1.733)=0.104$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic

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

$t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6$

P-value

First w eneed to find the degrees of freedom given by:

$df=n-1=16-1 =15$

Since is a two-sided test the p value would given by:

$p_v =2*P(t_{15}>2.6)=0.0201$

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, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0

3 years ago