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3 years ago

At a corner gas station, the revenue R varies directly with the number g of gallons of gasoline sold. If the revenue is

1 answer:

7 0

<span>At a corner gas station, the revenue R varies directly with the number g of gallons of gasoline sold. If the revenue is $44.50 when the number of gallons sold is 10, find a linear equation that relates revenue R to the number g of gallons of gasoline. Then find the revenue R when the number of gallons of gasoline sold is 15.5.

Solution:
As the question mentioned the direct relationship between the quantities, hence
10 gallons of gasoline sold = $44.50
15.5 gallons of gasoline sold = $x
by cross multiplication, we get that
10x = 15.5 * 44.50
which implies that
x = 68.975
Thus by $</span>68.975 revenue is obtained by selling 15.5 gallons of gasoline.

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Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

3 years ago

Answer this question:What is the difference between 9 < x and 9 = x? Explain.

maw [93]

9 < x

this means that x is a larger number than 9

9 = x

this means that x is equal to 9

3 0

3 years ago

See attachment for problem

Effectus [21]

The liters in the tank when it is filled to a height of 3.70 is 5,580 liters

The liters that needs to be added to 100% capacity is 480 liters

<h3>What is the volume?</h3>

A right circular cone is a three dimensional object has a flat circular base that tapers to a vertex. The volume of a right circular cone is the amount of space in the right circular cone.

Volume of a cone = 1/3(πr²h)

Where:

π = pi = 3.14
r = radius
h = height

Volume of the right circular cone when its filled to a height of 3.70 = 1/3 x 3.14 x 3.70 x 1.20² = 5.58 m³

5.58 x 1000 = 5,580 liters

Volume of the right circular cone when it is full = 1/3 x 3.14 x 4 x 1.20² = 6.03 m³

6.03 x 1000 = 6030 liters

Liters that needs to be added to 100% capacity = 6030 liters - 5,580 liters = 480 liters

To learn more about the volume of a cone, please check: brainly.com/question/13705125

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1 year ago

How to find the rate of change of the volume of cone?

sleet_krkn [62]

You can use calculus ( related rates). Given the rate os change of the radius for example you can find the rate of change of volume using differential calculus.

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3 years ago

Simplify.

natima [27]

Its 2y+7z hope dis helps

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3 years ago

Read 2 more answers

At a corner gas​ station, the revenue R varies directly with the number g of gallons of gasoline sold. If the revenue is

At a corner gas station, the revenue R varies directly with the number g of gallons of gasoline sold. If the revenue is