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3 years ago

What is the volume of the sphere in terms of pie

Mathematics

1 answer:

NeTakaya3 years ago

6 0

Answer: The volume is 288π .

Step-by-step explanation:

V= 4/3 *n*r^3

V = 4/3*n * 6^3

v=4/3 *n * 216

v= 288n

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Tell me pls but pls no files

Ilya [14]

Answer:

I think its A

Step-by-step explanation:

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2 years ago

It’s number 9 <br> I think it’s 5 or 15 IDK plz tell me the answer to A and B plz

Oduvanchick [21]

Answer:

you have it right its 29 and 5

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP RIGHT ANSWER GETS BRAINLIST

Elis [28]

Answer:

Step-by-step explanation:

Perimeter =5s+5s+2s+2s+0.75+0.75

2 off each implies part E

4 0

3 years ago

A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier bel

adelina 88 [10]

Answer:

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.

Step-by-step explanation:

1) Data given and notation

$\bar X=19.5$ represent the battery life sample mean

$s=1.9$ represent the sample standard deviation

$n=33$ sample size

$\mu_o =20$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

Since we don't know the population deviation, 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".

3) Calculate the statistic

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

$t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=33-1=32$

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

$p_v =P(t_{(32)}$

5) 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 average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours. TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

5 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago