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Pavel [41]
4 years ago
10

The following situation can be modeled by a linear function. Write an equation for the linear function and use it to answer the

given question. The price of a particular model car is ​$17, 000 today and rises with time at a constant rate of ​$820 per year. How much will a new car cost in 3.3 ​years?
Mathematics
1 answer:
Alenkinab [10]4 years ago
6 0

Answer:

The function is y = 17000 + 820 x

where: y = price of the car.

x = time in years.

After 3.3 years, the car will cost $19706.

Step-by-step explanation:

Hi there!

Let y be the price of the car.

After the first year, the price of the car will be:

y = 17000 + 820

After the second year:

y = 17000 + 820 + 820 or 17000 + 820(2)

After the third year:

y = 17000 + 820 + 820 + 820 or 17000 + 820(3)

Then, after x years:

y = 17000 + 820 x

After 3.3 year, the car will cost:

y = 17000 + 820(3.3) = 19706

After 3.3 years, the car will cost $19706.

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Please Help ME!!
ludmilkaskok [199]

A) 4*c + 2*D = 78

B) C + D =27

Multiplying B) by -2 equals

B) -2 C -2 D = -54 then adding equation A)

A) 4*c + 2*D = 78

2C = 24

Cows = 12

Ducks = 15

********************************** DOUBLE CHECK

B) C + D =27

12 + 15 = 27

Correct!

8 0
3 years ago
n a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of inches and a stand
kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

<u>The complete question is:</u> In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts​ (a) through​ (d) below. ​(a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = <u><em>the heights of men in the​ 20-29 age group</em></u>

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

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

where, \mu = population mean height = 67.5 inches

            \sigma = standard deviation = 2 inches

So, X ~ Normal(\mu=67.5, \sigma^{2}=2^{2})

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

 

      P(X < 67 inches) = P( \frac{X-\mu}{\sigma} < \frac{67-67.5}{2} ) = P(Z < -0.25) = 1 - P(Z \leq 0.25)

                                                                 = 1 - 0.5987 = <u>0.4013</u>

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

    P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X \leq 67 inches)

    P(X < 71 inches) = P( \frac{X-\mu}{\sigma} < \frac{71-67.5}{2} ) = P(Z < 1.75) = 0.9599

    P(X \leq 67 inches) = P( \frac{X-\mu}{\sigma} \leq \frac{67-67.5}{2} ) = P(Z \leq -0.25) = 1 - P(Z < 0.25)

                                                                = 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = <u>0.5586</u>.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

 

      P(X > 71 inches) = P( \frac{X-\mu}{\sigma} > \frac{71-67.5}{2} ) = P(Z > 1.75) = 1 - P(Z \leq 1.75)

                                                                 = 1 - 0.9599 = <u>0.0401</u>

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

7 0
3 years ago
Factorise 12x cubed - 9x squared
TiliK225 [7]

Answer:

3x²(4x-3)

Step by step explanation:

Factorise 12x³-9x²

first, find the HCF of the terms.

the HCF is 3x²

Then you can use the HCF to multiply all terms, each divided by the HCF.

3x²(12x³/3x²-9x²/3x²)

Dividing, we'll have

3x²(4x-3)

6 0
3 years ago
Help wanted pleasse. I'd be very thankful, you'd be even more awesome
Sladkaya [172]
Whay kind of math is it
6 0
3 years ago
If f(3) = 9, what is f(1)<br>​
Vaselesa [24]

Answer:

3

Step-by-step explanation:

3*3=9

1*3=3

8 0
3 years ago
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