Find the volume of a cube if each side is 3.3 inches long. Do not round your answer.

Let f(x)=x^2+2x+3 . What is the average rate of change for the quadratic function from x=−2 to x = 5?

miskamm [114]

To find the average rate of change of given function f(x) on a given interval (a,b):

Find f(b)-f(a), b-a, and then divide your result for f(b)-f(a) by your result for b-a:

f(b) - f(a)
------------
b-a

Here your function is f(x) = x^2 - 2x + 3. Substituting b=5 and a=-2,
f(5) = 5^2 -2(5)+3 =? and f(-2) = (-2)^2 - 2(-2) + 3 = ?

Calculate f(5) - [ f(-2) ]
------------------ using your results, above.
5 - [-2]

Your answer to this, if done correctly, is the "average rate of change of the function f(x) = x^2+2x+3 on the interval [-2,5]."

5 0

3 years ago

How many people live in the state of Tennessee?

DochEvi [55]

6.549 million in 2014 i hope this helps.

8 0

3 years ago

Read 2 more answers

Suppose y=f(x)+k. what effect does k have on the parent function?

Aleksandr [31]

It moves it up and down

6 0

3 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we 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:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

5 0

2 years ago

Suppose we describe the weather as either hot (Upper H) or cloudy (Upper C). Answer parts (a) and (b) below.

Ket [755]

Answer:

{HH,CC,HC,CH}

Step-by-step explanation:

We are given that

H denotes hot and cloudy denotes C.

We have to find the total possible outcomes for the weather on two consecutive days.

The possible cases in two consecutive days

Both days are hot=HH

Both days are cloud=CC

First day is hot other day cloudy=HC

First day is cloudy other day is hot=CH

Total possible cases=HH,CC,HC,CH

Therefor, the total outcomes for the weather on two consecutive days={HH,CC,HC,CH}

4 0

3 years ago