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Goryan [66]
4 years ago
15

What is the average of 143, 98, and 128

Mathematics
1 answer:
asambeis [7]4 years ago
8 0
Average\ A=\frac{S}{N}\\\\A=\frac{143+98+128}{3}=\frac{369}{3}=123\\\\Average\ of\ these\ numbers\ is\ 123.
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How does the volume of a sphere compare to that of a cylinder with a height and radius equal to the radius of the sphere?
disa [49]

1) The height of the cylinder is equal to the diameter of the sphere.

3) The radius of the sphere is half the height of the cylinder.

5) The volume of the sphere is two-thirds the volume of the cylinder.

Hope this helped! :)

8 0
3 years ago
Tim's bank contains quarters, dimes and nickels. He has 3 more dimes than quarters and 6 fewer nickels than quarters. If he has
Artist 52 [7]
(x - 3) + (x - 6) + x = 63
x - 3 + x - 6 + x = 63
Combine like terms
3x - 9 = 63
Isolate the constant
3x - 9 + 9 = 63 + 9
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Isolate the viable
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8 0
4 years ago
What is 22% of 64.82
Phantasy [73]

Answer:

14.2604

Step-by-step explanation:

this is because 22% as a decimal would be 0.22, and since this is out of 64.82, you have to multiply 0.22 by 68.82

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Hope this helps <33

4 0
3 years ago
Read 2 more answers
20 less than the product of 5 and x
baherus [9]

Answer:

5x-20

Step-by-step explanation:

since it says "product of 5 and x" we know that 5 and x is being multiplied

5x

since it says 20 less than we know that we need to subtract 20 from 5x

5x-20

5 0
3 years ago
Read 2 more answers
A certain population of mice is growing exponentially. After one month there are 48. After 2 months there are "192" mice. Write
REY [17]

Answer:

P(t) = 12e^1.3863k

Step-by-step explanation:

The general exponential equation is represented as;

P(t) = P0e^kt

P(t) is the population of the mice after t years

k is the constant

P0 is the initial population of the mice

t is the time in months

If after one month there are 48 population, then;

P(1) = P0e^k(1)

48 = P0e^k ...... 1

Also if after 2 months there are "192" mice, then;

192 = P0e^2k.... 2

Divide equation 2 by 1;

192/48 = P0e^2k/P0e^k

4 = e^2k-k

4 = e^k

Apply ln to both sides

ln4 = lne^k

k = ln4

k = 1.3863

Substitute e^k into equation 1 to get P0

From 1, 48 = P0e^k

48 = 4P0

P0 = 48/4

P0 = 12

Get the required equation by substituting k = 1.3863 and P0 = 12 into equation 1, we have;

P(t) = 12e^1.3863k

This gives the equation representing the scenario

8 0
3 years ago
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