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Snowcat [4.5K]
3 years ago
5

Each cracker in the stack has a diameter of 2 inches. The height of the stack is 7.5 inches. What is the volume of the stack?

Mathematics
1 answer:
otez555 [7]3 years ago
7 0
THe stack forms a cylinder.

Volume = pi r^2 h   

r = 1/2*2 = 1 inch, height = 7.5 inches

Volume of stack =  pi*1^2* 7.5 =  23.56 ins^3  Answer
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-1/3 + (-7/4)<br><br> Ill give this cookie to who solves it
elena-14-01-66 [18.8K]

Answer:

-25/12

Step-by-step explanation:

first, you need a common denominator. in this case, it's 12.

-1/3=-4/12

-7/4=-21/12

and then you add

-4+-21=-25

so -25/12 is your answer

5 0
2 years ago
Read 2 more answers
work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
Yuki888 [10]

Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the​ conference is 0.10.

(c) The probability that Clarice attends the​ conference is 0.40.

(d) The probability that John stays​ home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (<em>M</em>), Roberto (<em>R</em>), John (<em>J</em>), Dominique (<em>D</em>) and Clarice (<em>C</em>) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

{5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

The probability of the event <em>E</em> is:

P(E)=\frac{n(E)}{N}

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of  Roberto and John is, <em>RJ</em>.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the​ conference as follows:

P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10

Thus, the probability that Roberto and John attend the​ conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the​ conference as follows:

P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40

Thus, the probability that Clarice attends the​ conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays​ home as follows:

P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60

Thus, the probability that John stays​ home is 0.60.

4 0
3 years ago
PLEASE ANSWER ASAP!!!
qwelly [4]

Answer:

D, E

Step-by-step explanation:

It’s not A because the equation can not be any pair of numbers.

It’s not B because that is stupid logic.

It’s not C because (3,4) and (10,8) don’t work

It’s not F because there is 1 solution

5 0
3 years ago
Estimate 34,926 - 17,042
Ann [662]
I think the anwser is 17,887
7 0
3 years ago
Kathy takes her cat to a veterinarian every year for a check-up. Last year, the difference in the cat's weight from teh year bef
Nataliya [291]
To find the difference in weights from two years ago, you will need to combine the 2 differences from the two years.  

-1.56 and 0.73 becomes -1.56 + 0.73.

To add a positive and a negative number you will subtract the absolute values of the numbers AND use the sign of the number with the bigger absolute value (negative) in your answer.

|-1.56| - |0.73|
1.56 - 0.73 = 0.83
-0.83

The cat lost 0.83 (-0.83) pounds over 2 years.
3 0
3 years ago
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