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pogonyaev
2 years ago
14

What is the sum or whatever of this equation. This equation is V = πx3^2x8

Mathematics
1 answer:
navik [9.2K]2 years ago
6 0

Answer:

V=226.08

Step-by-step explanation:

To evaluate V from the following expression:

V=\pi\times3^2\times 8

Taking \pi=3.14

V=3.14\times 3^2\times 8

V=3.14\times 9\times 8

∴ V=226.08

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The value of 110 coins, consisting of dimes and quarters, is $20.30. how many of each kind of coin are there?
deff fn [24]
D + q = 110......d = 110 - q
0.10d + 0.25q = 20.30

0.10(110 - q) + 0.25q = 20.30
11 - 0.10q + 0.25q = 20.30
-0.10q + 0.25q = 20.30 - 11
0.15q = 9.30
q = 9.30/0.15
q = 62 <==== there are 62 quarters

d = 110 - q
d = 110 - 62
d = 48 <==== there are 48 dimes

6 0
3 years ago
(Pls help) A pie crust recipe calls for 2 cups of butter for each 3 cups of flour. Ella is making enough
blagie [28]

.6 cups of flour Step-by-step explanation:

4 0
2 years ago
A town has a population of 50,000. it’s rate increase is 8% every 6 months. find the population after 4 years
Ainat [17]
82,000 bc 8% of 50,000 is 4000 and 4000 times 8 is 32,000 + 50000
6 0
2 years ago
Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule
Vaselesa [24]

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

<em>a) What is the approximate percentage of men between 169 and 183 cm? </em>

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

<em>b) Between which 2 heights would 95% of men fall?</em>

This corresponds to ±2 standard deviations off the mean.

\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190

95% of the men will fall between 162 cm and 190 cm.

<em>c) Is it unusual for a man to be more than 197 cm tall?</em>

The number of standard deviations of distance from the mean is

n=(197-176)/7=3

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

3 0
3 years ago
PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE HELP ME!!!!!!!! THATS ALL. I WILL GIVE POINTS, FAN, MEDAL. PLEASE, I AM SO BAD AT MATH
aleksandrvk [35]
"measures to the nearest tenth of a pound" tell us to round each of the weights listed to the nearest tenth (since the unit is already pounds, no conversion is needed). What this means is if the second number to the right of the decimal is 5 or greater, then we add 1 to the first number to the right of the decimal (and replace the second number with a zero). With all this said, it's much easier to show you, rather than to try to explain it. :D

4.35 is rounded up to 4.4 because the second number to the right of the decimal is 5.
Packet A = 4.4
Packet B = 4.4 
Packet C = 4.5
Packed D = 4.4
The question is asking what the heaviest reading will be, so from the whole list above we pick the highest number which is 4.5 pounds. Notice how the answer is not Packet C, or 4.48 pounds, but 4.5 pounds. It is asking for the reading, not the packet, nor the weight. Hope this helps. 
7 0
3 years ago
Read 2 more answers
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