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3 years ago

If the ratio of the triangular pyramids' surface areas is 4:9, what is F?

Mathematics

1 answer:

VladimirAG [237]3 years ago

4 0

Answer:

The answer to your question is f = 10.35 in

Step-by-step explanation:

Data

Proportion = 4:9

height 1 = 4.6 in

height 2 = f

Process

1.- Write the proportion

$\frac{4}{4.5} = \frac{9}{f}$

2.- Solve for f

$f(\frac{4}{4.6}) = f(\frac{9}{f})$

$f(\frac{4}{4.6}) = 9$

$4.6\frac{4f}{4.6} = 9(4.6)$

4f = 41.4

$\frac{4f}{4} = \frac{41.4}{4}$

f = 10.35 in

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Roopesh has $24 dollars to spend on a birthday gift. The store where he is shopping has a sale offering $5 off the regular price

Natasha2012 [34]

Answer:

r ≤ 29, r-5

The sale price can be compared with the regular price, r-5 ≤ 24

Step-by-step explanation:

Amount to spend = $24

Regular price = r

Sale = $5

Sale Price = r-5

The regular price will be $5, at the max, more than the amount Roopesh has to spend.

The sale price will be $24 or less than that for Roopesh to afford.

Inequality for regular price:

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3 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$