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3 years ago

URGENT WILL GET 20 and Brainlest

1 answer:

8 0

The polynomial that represents the difference between the cost of distributing by train and the cost of distributing by truck is:

$f(x) = -0.04x^2 + 18x + 75$

Solution:

The cost of distributing by train can be modeled as:

$-0.07x^2 + 40x - 105$

The cost of distributing by truck can be modeled as:

$-0.03x^2 + 22x - 180$

where x is the number of tons of product distributed

To find: Difference between the cost of distributing by train and the cost of distributing by truck

Difference = cost of distributing by train - cost of distributing by truck

Let f(x) = difference in cost between trains & trucks

$f(x) = (-0.07x^2+40x-105) - (-0.03x^2 +22x - 180)$

$f(x) = -0.07x^2 + 40x -105 + 0.03x^2 -22x + 180$

$Combine\ the\ like\ terms\\\\f(x) = -0.07x^2 +0.03x^2 + 40x - 22x +180-105\\\\Add\ the\ like\ terms\\\\f(x) = -0.04x^2 + 18x + 75$

Thus polynomial that represents the difference between the cost of distributing by train and the cost of distributing by truck is found

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3 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

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