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3 years ago

Look at the formula for the volume of a cylinder below.

1 answer:

8 0

$h = V / ( \pi r^2)$

Okay, so we already have the volume equation for a cylinder here.
$V = \pi r^2 h$
Since we need to find the height(h), we need to basically isolate it in the equation. To do that we just have to divide both sides by (pi * r^2), where we end up with:
$V / (\pi r^2) = h(\pi r^2) / (\pi r^2)$
then
$V / ( \pi r^2) = h$
$h = V / ( \pi r^2)$

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How can I remember the special right triangle formulas?

lara31 [8.8K]

Maybe you can study it or figure out a trick. I usually study with my sibling, but if you don't have any, friends or other family. I am sure they won't mind. Also, maybe making flashcards, those work well.
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4 years ago

A translation maps (x, y) to

Troyanec [42]

Answer:

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Step-by-step explanation:

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3 years ago

Which value is equivalent to cos 10°?<br> sin 30°<br> O sin 25°<br> O sin 70°<br> O sin 80°

Sauron [17]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

1+1+1+1 what is the correct answer

Semenov [28]

1+1+1+1

First, let's do 1+1

1+1=2

and the other 1+1 also equals 2

2+2=4

so the final answer is 4

5 0

2 years ago

Read 2 more answers

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

$\ln{e^{-0.01x}} = \ln{0.08}$

$-0.01x = \ln{0.08}$

$x = -\frac{\ln{0.08}}{0.01}$

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago