Mathematics, 04.11.2020 18:10 cicilee49

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

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Answer from: hannahgracew12

Complete Question:

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

a. Every car selected is in poor shape

b. At least two cars selected are in good shape.

c. Exactly three cars selected are in great shape.

d. Two cars selected are in great shape and two are in good shape.

e. One car selected is in good shape but the other 3 selected are in poor shape.

$P_A = \frac{^4 C_4}{^{30}C_{4}}$

$P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}$

$P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}$

$P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}$

$P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}$

Step-by-step explanation:

From the question we are told that

The number of car in the parking lot is n = 30

The number of cars in great shape is k = 10

The number of cars in good shape is r = 16

The number of cars in poor shape is q = 4

The number of cars that were selected at random is N= 4

Considering question a

Generally the number of way of selecting four cars that are in a poor shape from number of cars that are in poor shape is

$^{q} C_{N}$

=> $^{4} C_{4}$

Here C stands for combination.

Generally the number of way of selecting four cars that are in a poor shape from total number of cars in the parking lot is

$^{n} C_{N}$

=> $^{30} C_{4}$

Generally the probability that every car selected is in poor shape is mathematically represented as

$P_A = \frac{^4 C_4}{^{30}C_{4}}$

Considering question b

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

$^{r} C_{2}$

=> $^{16} C_{2}$

Here C stands for combination.

Generally the number of way of selecting the remaining 2 cars from the remaining number of cars in the parking lot is

$^{n-r} C_{2}$

=> $^{30-16} C_{2}$

=> $^{14} C_{2}$

Generally the number of way of selecting 3 cars that are in good shape from number of cars that are in good shape is

$^{r} C_{3}$

=> $^{16} C_{3}$

Generally the number of way of selecting the remaining 1 cars from the remaining number of cars in the parking lot is

$^{n-r} C_{1}$

=> $^{30-16} C_{1}$

=> $^{14} C_{1}$

Generally the number of way of selecting 4 cars that are in good shape from number of cars that are in good shape is

$^{r} C_{4}$

=> $^{16} C_{4}$

Generally the probability that at least two cars selected are in good shape

$P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}$

Considering question c

Generally the number of way of selecting 3 cars that are in great shape from number of cars that are in great shape is

$^{k} C_{3}$

=> $^{10} C_{3}$

Generally the number of way of selecting the remaining 1 cars from the remaining number of cars in the parking lot is

$^{n-k} C_{1}$

=> $^{30-10} C_{1}$

=> $^{20} C_{1}$

Generally the probability of selecting exactly three cars selected are in great shape is

$P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}$

Considering question d

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

$^{r} C_{2}$

=> $^{16} C_{2}$

Generally the number of way of selecting 2 cars that are in great shape from number of cars that are in great shape is

$^{k} C_{2}$

=> $^{10} C_{2}$

Generally the probability that two cars selected are in great shape and two are in good shape.

$P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}$

Considering question e

Generally the number of way of selecting 1 cars that is in good shape from number of cars that are in good shape is

$^{r} C_{1}$

=> $^{16} C_{1}$

Generally the number of way of selecting 3 cars that are in a poor shape from number of cars that are in poor shape is

$^{q} C_{3}$

=> $^{4} C_{3}$

Generally the probability that one car selected is in good shape but the other 3 selected are in poor shape is

$P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}$

Answer from: Quest

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

i hope this !

Answer from: Quest

the answer is 489

step-by-step explanation:

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