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Ivahew [28]
3 years ago
8

Profit, P(x), is the difference between revenue, R(x), and cost, C(x), so P(x) = R(x) - CC). Which

Mathematics
2 answers:
aleksklad [387]3 years ago
7 0

Answer:

Step-by-step explanation:

Profit, P(x), is the difference between revenue, R(x), and cost, C(x)

R(x) = 2x^4 - 3x^3 + 2x - 1

C(x) = x^4 -x^2 + 2x + 3

Substituting into the equation

P(x) = R(x) - C(x)

= (2x^4 - 3x^3 + 2x - 1) - (x^4 -x^2 + 2x + 3)

= x^4 - 3x^3 + x^2 - 4

The answer is the lower left option.

JulsSmile [24]3 years ago
3 0

Answer:

P(x)=x^4-3x^3+x^2-4

(This is the option found in the lower-left corner)

Step-by-step explanation:

When given the following functions,

R(x)=2x^4-3x^3+2x-1

C(x)=x^4-x^2+2x+3

The problem asks one to find (P(x)), moreover, one is given the following information,

(P(x))=(R(x))-(C(x))

Substitute,

P(x)=(2x^4-3x^3+2x-1)-(x^4-x^2+2x+3)

Simplify, multiply everything in the second parenthesis by the negative sign outside of it,

P(x)=2x^4-3x^3+2x-1-x^4+x^2-2x-3

Combine like terms, only operations between coefficients of the same variable with the same degree (exponent) can be performed,

P(x)=x^4-3x^3+x^2-4

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let D^c be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,P(D)=\frac{1}{10000}

Only 0.4% of the people who don't have it test positive.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

P(D^c)=1-\frac{1}{10000}

Formula:P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}

P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

(B)

P(D^c|A^c)=probability that someone does not have disease given that he tests negative

P(A^c|D^c)=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

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Step-by-step explanation:

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