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Reika [66]
3 years ago
14

A cereal box has dimensions of 1012 inches by 712 inches by 212

Mathematics
2 answers:
STatiana [176]3 years ago
6 0

Answer - 196.875 in3

(This will be the correct answer)

Juliette [100K]3 years ago
3 0
The answer is 196.875 in3

<span>Assuming you meant the following measures

10 1/2 inches by 7 1/2 inches by 2 1/2 inches.

The volume is V = L*W*H = 10.5 in * 7.5 in *2.5 in = 196.875 in3</span>
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Peggy had three times as many quarters as nickels. She had $1.60 in all. How many nickels and how many quarters did she have? Wh
svlad2 [7]
Let q and n represent the number of quarters and nickels respectively.

q=3n and .05n+.25q=1.6  These are the conditions in mathematical terms.

To solve, use the value of q from the first equation in the second equation to get:

.05n+.25(3n)=1.6  carry out indicated multiplication on left side

.05n+.75n=1.6  combine like terms on left side

.8n=1.6  divide both sides by .8

n=2, since q=3n

q=6

So Peggy had 2 nickels and 6 quarters.
3 0
3 years ago
A vending machine operator has determined that the number of candy bars sold per week by a certain machine is a random variable
docker41 [41]

Answer:

The answer is below

Step-by-step explanation:

Let Y represent the profit per day, and x represent the number of bar sold per day. Hence:

Y = 0.25x - 2

a) The mean is given as:

\mu_y=E(Y)=E(0.25X-2) = 0.25E(X)-2\\\\But\ E(X)=mean\ of\ machine=125. Hence:\\\\E(Y)=0.25(125)-2=29.25\\\\E(Y)=\$29.25

b) The standard deviation of y is:

\sigma_y=var(0.25X-2)\\\\\sigma_y=0.25\sigma(X)\\\\but\ \sigma_X=7,hence:\\\\\sigma_y=0.25(7)=1.75\\\\\sigma_y=\$1.75

6 0
3 years ago
Fernando and Roger are both in an art class. Fernando has created 40 projects in class this year. Fernando has created five time
Natali [406]
40x5= 200 I'm not sure who created five times more projects than the other, but whoever did, created 200 projects.
3 0
3 years ago
A circle has a diameter of 8m. What is its circumference?
Maslowich
The answer should be 25.12
6 0
3 years ago
what is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube
Dmitrij [34]
Let the least possible value of the smallest of 99 cosecutive integers be x and let the number whose cube is the sum be p, then

\frac{99}{2} (2x+98)=p^3 \\  \\ 99x+4,851=p^3\\ \\ \Rightarrow x=\frac{p^3-4,851}{99}

By substitution, we have that p=33 and x=314.

Therefore, <span>the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube is 314.</span>
3 0
3 years ago
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