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Nataliya [291]
3 years ago
11

!!!HELP!!! PLEASE SHOW WORK SO I CAN UNDERSTAND!

Mathematics
1 answer:
ss7ja [257]3 years ago
6 0

The only part of this problem that's the least bit tricky is knowing
the formula for the volume of a sphere.  Here is is:

          Volume of a sphere = (4/3) (pi) (radius of the sphere)³  .

Everything else is just arithmetic.

pi . . . use 3.14
Radius . . . 2-1/8 inches.  It'll be easier if we write that as  2.125 inches.

Now, just put the numbers into their places in the formula:

         Volume = (4/3) (pi) (radius of the sphere)³

                     = (4/3) (3.14) (2.125)³

and run that through your calculator.

When I run it through mine, I get  <u>40.174 in³ .</u>

None of the choices is anywhere near that number.
I'm worried, and I don't understand that.

Maybe we should look back at the question, and notice that
it says "estimate" ... something you do quickly in your head.

OK.  We know the real formula ...   Volume = (4/3) (pi) (radius)³ .
How could we estimate this quickly in our head?

pi . . . instead of 3.14, just use 3

radius . . . instead of 2-1/8, just use 2 .  Then (radius)³ = 2 x 2 x 2 = 8

Then  (4/3) (pi) (radius)³ = (4/3) (3) (8) = (4/3) of 24  =  <em>32 .</em>

Well now, will you look at that !  I guess that's where the 32 came from
on the list of choices.


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Read 2 more answers
A professor believes the class he is teaching in the current semester is above the average of the classes he has taught. In orde
MakcuM [25]

Answer:

1) Since we know the info from all the students that he teaches and we know the population deviation from past data we can use a z test to check the hypothesis

2) t=\frac{82-78}{\frac{15}{\sqrt{120}}}=2.921    

3) p_v =P(Z>2.921)=0.0017  

Since the p value is lower than the significance level of 0.01 we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis

4) For this case since we reject the null hypothesis we have enough evidence ot conclude that the scores for this semester are above the historical value of 78 so then the claim stated by the teacher makes sense

Step-by-step explanation:

Part 1

Since we know the info from all the students that he teaches and we know the population deviation from past data we can use a z test to check the hypothesis

Part 2

\bar X=82 represent the sample mean  for the scores

\sigma=15 represent the population standard deviation

n=120 represent the sample selected

\alpha=0.01 significance level  

System of hypothesis

He wants to test if the group for this current semester is above the average of the classes he has taught (mean 78), the system of hypothesis are:

Null hypothesis:\mu \leq 78  

Alternative hypothesis:\mu > 78  

Since we know the population deviation we can use the following statistic

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

Replacing the info given we got:

t=\frac{82-78}{\frac{15}{\sqrt{120}}}=2.921    

Part 3

We can calculate the p value for this test with this probability taking in count the alternative hypothesis:

p_v =P(Z>2.921)=0.0017  

Since the p value is lower than the significance level of 0.01 we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis

Part 4

For this case since we reject the null hypothesis we have enough evidence ot conclude that the scores for this semester are above the historical value of 78 so then the claim stated by the teacher makes sense

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Evaluate m+n2 if we know m=2 and n=-2 A.-6 B.8 C.-8 D.6
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The answer would be -2 when using orders of operation because 2x-2 is -4+2=-2
5 0
3 years ago
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