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Dmitry [639]
2 years ago
13

The starting salaries of individuals with an mba degree are normally distributed with a mean of $40,000 and a standard deviation

of $5,000. refer to exhibit 6-4. what is the probability that a randomly selected individual with an mba degree will get a starting salary of at least $30,000
Mathematics
1 answer:
horsena [70]2 years ago
6 0
<span>With the mean of 40k and standard deviation of 5k, we need to find P(x>=30k). P((30k-40k)/5k) = P(z<-2) = 1-0.0228 which is equal to 0.9772. There is a 97.72% chances that the starting salary will be at least 30k.</span>
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43.20

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Is the line y = 3x – 7 parallel or perpendicular to 3x + 9y = 9? Explain your answer.
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2 years ago
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On a certain​ route, an airline carries 7000 passengers per​ month, each paying ​$30. A market survey indicates that for each​ $
KengaRu [80]

Answer:

The ticket price that maximizes revenue is $50.

The maximum monthly revenue is $250,000.

Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

R=p\cdot N=(30+x)(7000-x)

where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

p=30+x=30+20=50

The maximum monthly revenue is:

R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000

3 0
2 years ago
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denis-greek [22]

Given:

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8 0
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4 0
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