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3 years ago

The equation V=13πr2h relates the volume V, the radius r, and the height h of a cone. Write h in terms of V and r.

Mathematics

2 answers:

maksim [4K]3 years ago

6 0

$V= \frac{1}{3} \pi r^2h \\ \\ 3V= \pi r^2h \\ \\ h= \cfrac{3V}{\pi r^2}$

Gnom [1K]3 years ago

5 0

V=(1/3)pir^2h
times both sides by 3
3V=pir^2h
divide both sides by (pir^2)
$\frac{3V}{\pi r^2}=h$

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When Rafael emptied his pockets, e found he had a total of $3.50 in quarters and nickels. If he had 8 more quarters that nickels

jasenka [17]

You can use systems of equations for this one.

We are going to use 'q' as the number of quarters Rafael had,
and 'n' as the number of nickels Rafael had.

You can write the first equation like this:
3.50=0.05n+0.25q
This says that however many 5 cent nickels he had, and however many
25 cent quarters he had, all added up to value $3.50.
Our second equation is this:
q=n+8
This says that Rafael had 8 more nickels that he had quarters.

We can now use substitution to solve our system.

We can rewrite our first equation from:
3.50=0.05n+0.25q
to:
3.50=0.05n+0.25(n+8)

From here, simply solve using PEMDAS.

3.50=0.05n+0.25(n+8)     --Distribute 0.25 to the n and the 8
3.50=0.05n+0.25n+2     --Subtract 2 from both sides
1.50=0.05n+0.25n     --Combine like terms
1.50=0.30n     --Divide both sides by 0.30
5=n     --This is how many NICKELS Rafael has.

We now know how many nickels he has, but the question is asking us
how many quarters he has.

Simply substitute our now-known value of n into either of our previous
equations (3.50=0.05n+0.25q or q=n+8) and solve.

We now know that Rafael had 13 quarters.

To check, just substitute our known values for our variables and solve.
If both sides of our equations are equal, then you know that you have
yourself a correct answer.

Happy math-ing :)

5 0

3 years ago

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

7 0

3 years ago

Which whole number is closest to 577.41 closest to?

Stella [2.4K]

578 bc 2 goes into 578 evenly

7 0

4 years ago

a block of gold with a mass of 4,830 kilograms has a volume of 0.25 cubic meters. what is the density of the gold?

GrogVix [38]

Density = mass / volume
Density = 4,830 / 0.25
Density = 19,320 kg/m

8 0

4 years ago

Why is 751, 447 rounded to 800,000

disa [49]

I believe its because anything above 5 is rounded up, 7 rounded up is 8 and how I was taught is if its a number (this case its 751,447) the 7 in it would round to 8 causing all the other numbers to go up as well causing the 800,000 >.< I

5 0

4 years ago

Read 2 more answers