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Anuta_ua [19.1K]
3 years ago
15

Maria just put 23.38 gallons of fuel into her car. There were 1.7 gallons in the car to begin with. How much fuel is in Maria's

car now?
Mathematics
1 answer:
natulia [17]3 years ago
6 0

Answer:

25.08

Step-by-step explanation:

I just added 1.7+23.38

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A small publishing company is releasing a new book. The production costs will include a one-time fixed cost for editing and an a
svetoff [14.1K]

The profit equation which relates P, profit to N, quantity produced and sold is:

P=17.75N-750

What is the total cost function?

The total production cost function, whereby total cost is C is as given below:

C=750+15.95N(look at a similar question provided)

What is the total revenue function?

The total revenue function is as given below:

R=33.70N

The profit is the excess of total revenue over total cost, in other words, profit is the total revenue minus total cost

P=33.70N-(750+15.95N)

P=33.70N-750-15.95N

P=33.70N-15.95N-750

P=17.75N-750

Find out more about profit function on: brainly.com/question/25668243

#SPJ1

Similar question:

A small publishing company is releasing a new book. The production costs will include a one-time fixed cost for editing and an additional cost for each book printed. The total production cost C (in dollars) is given by the function C=650+18.95N , where N is the number of books. The total revenue earned (in dollars) from selling the books is given by the function R=32.80N . Let P be the profit made (in dollars). Write an equation relating P to N . Simplify your answer as much as possible.

4 0
2 years ago
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
4 years ago
K-10=-24 what is the value of k
pickupchik [31]

Answer:

14

Step-by-step explanation:

k = 10 + - 24

k = -14

8 0
4 years ago
Read 2 more answers
1/2=2m multiplication and division equations
krek1111 [17]
1/2 = 2m

2m/2 = 1/2 divided by 2

1/2 * 1/2 = 1/4

m = 1/4
m = 0.25

Answer: m = 1/4 or 0.25
3 0
3 years ago
Hey can anyone pls asnwer dis!
VashaNatasha [74]

Answer:

a

Step-by-step explanation:

5 0
3 years ago
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