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kap26 [50]
3 years ago
10

The volume of the cylinder

Mathematics
2 answers:
sattari [20]3 years ago
5 0

Answer:

15550.88

Step-by-step explanation:

the formula for the volume of a cylinder is πr^2(h), so plug in height and radius, and you have the answer.

Drupady [299]3 years ago
5 0

Answer:

15550.88

Step-by-step explanation:

You use πr^2 h

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An airplane manufacturer needs to purchase an aluminum alloy for use in constructing a part for their new line of jumbo jets. It
Nostrana [21]

Answer:

Step-by-step explanation:

Hello!

In this example, you have 3 populations determined by each aluminum alloy A, B, and C and want to test if their tensile strength is the same or not. So the study variable is the tensile strength of the aluminum and it is tested in each sample taken of each alloy. The parameters of interest are the average tensile strength of each alloy.

The best statistical procedure to compare these three population is an ANOVA test of one factor(tensile strength) with three levels (alloy A, alloy B, and alloy C)

I hope it helps!

6 0
2 years ago
Answer the following questions.
Alona [7]
The answer for this one is $62.04
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2 years ago
How to solve -2(1-7n)-5n=34
OLEGan [10]

-2(1-7n)-5n=34

Multiply the bracket by -2

(-2)(1)=-2

(-2)(-7n)=14n

-2+14n-5n=34

-2+9n=34

Move -2 to the other side. Sign changes from -2 to +2.

-2+2+9n=34+2

9n=34+2

9n=36

Divide by 9 for both sides

9n/9=36/9

n=4

Answer: n=4

6 0
2 years ago
Read 2 more answers
You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white
Rom4ik [11]

Answer:

Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b: <em>The case in such a way that the chances are maximized so the case  where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c: <em>The minimum and maximum probabilities of winning  for n number of balls are  such that </em>

  • <em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>
  • <em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

so

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

so

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball  and 1 white ball in urn A and 2 white balls are in urn A'

so

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball  and 2 white balls in urn A and 1 white ball are in urn A'

so

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b:

<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c:

The minimum and maximum probabilities of winning  for n number of balls are  such that

  • when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

  • when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0
3 years ago
If you answered 37 problems correctly on a 42 question test, what percent would you recieve
OleMash [197]
The fraction would be 37 out of 42. You basically divide the two and get something around 88%.
5 0
3 years ago
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