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omeli [17]
3 years ago
10

A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil contained in 1 mile (5280 feet) of a pipeli

ne that has an inside diameter of 6 inches and is completely filled with oil? How much is "1 mile" of oil in this pipeline worth at a price of $100 per barrel?
Mathematics
1 answer:
8_murik_8 [283]3 years ago
6 0
1)volume of the  pipeline

The pipeline is a cylinder, therefore;
Volume (cylinder)=πr²h
r=radius
h=height of the cylinder

diameter=6 in*(1 ft / 12 in)=0.5 ft
raius=diameter / 2=0.5 ft / 2=0.25 ft.
height=5280 ft

Volume (pipeline)=π(0.25 ft)²(5280 ft)=330π ft³≈1036.73 ft³.


2) we calculate the number of barrel
1 mile of oil in this pipeline is 330π ft³ of oil.

1 barrel of crude------------------5.61 ft³
x----------------------------------330π ft³

x=(1 barrel*330π ft³) / 5.61 ft³=184.8 barrels

3) we calculate the price.

1 barrel---------------$100
184.8 barrels----------  x

x=(184.8 barrels * $100) / 1 barrel=$18,480

Solution: ≈$18,480
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a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that p = 0.25

25 incoming calls.

This means that n = 25

a. What is the probability that at most 4 of the calls involve a fax message?

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4). Then

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

P(X \leq 4) + P(X > 4) = 1

From a), P(X \leq 4) = 0.214)

Then

P(X > 4) = 1 - 0.214 = 0.786

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0
2 years ago
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