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Effectus [21]
3 years ago
15

An automotive repair center charges $45 for any part of the first hour of labor, and $25 for any part of each additional hour. W

hich of the following is a correct cost?
A. C(t) = 145 for 5 < x ≤ 6
B. C(t) = 145 for 6 < x ≤ 7
C. C(t) = 170 for 5 < x ≤ 6
D. C(t) = 170 for 6 < x ≤ 7
Mathematics
1 answer:
IceJOKER [234]3 years ago
7 0
The total cost is given by the equation:
C(t) = 45 + 25(h-1) where h is the number of hours worked.

We can check for each option in turn:

Option A:

Inequality 5 < x ≤ 6 means the hour is between 5 hours (not inclusive) to 6 hours (inclusive)
Let's take the number of hours = 5
C(5) = 45 + (5-1)×25 = 145
Let's take the number of hours = 6
Then substitute into C(6) = 45 + (6-1)×25 = 170
We can't take 145 because the value '5' was not inclusive.


Option B:
The inequality is 6 < x ≤ 7
We take number of hours = 6
C(6) = 25(6-1) + 45 = 170
We take number of hours = 7
Then C(7) = 25(7-1) + 45 = 195

Option C:
The inequality is 5 < x ≤ 6
Take the number of hours = 5
C(5) = 25(5-1) + 45 = 145
Take the number of hours = 6
C(6) = 25(6-1) + 45 = 170
We can't take the value 145 as '5' was not inclusive in the range, but we can take 170

Option D:
6 < x ≤ 7
25(6-1) + 45 < C(t) ≤ 25(7-1) + 45
170 < C(t) ≤ 195

Correct answer: C


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Answer:

A) probability the sum is 8 or 11= 4/21

B) probability that sum is 12 or less than 10 = 6/7

C) Probability that the sum is 3 or less than 3 = 2/21

D) Probability that the sum is 2 or 10 = 1/7

Step-by-step explanation:

Since we have the same probability of each event in each dice, the answer would be just to check the different outcomes, two dices, each with 1 to 6;

(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,2);(2,3);(2,4);(2,5);(2,6);(3,3);(3,4);(3,5);(3,6);(4,4);(4,5);(4,6);(5,5);(5,6);(6,6)

Thus, there are 21 possible outcomes.

Now,

A) probability that the sum is 8 or 11;

From the outcomes above, the number of outcomes that have a sum as 8 or 11 are;

(2,6) ; (3,5) ; (4,4) ; (5,6)

So,probability = 4/21

B) From the outcomes above, the number of outcomes that are 12 or less than 10 are;

(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,2);(2,3);(2,4);(2,5);(2,6);(3,3);(3,4);(3,5);(3,6);(4,4);(4,5);(6,6).

There are 18 possible outcomes.

So, probability that sum is 12 or less than 10 = 18/21 = 6/7

C)From the initial 21 outcomes, the number of outcomes that the sum is 3 or less than 3 are;(1,1);(1,2)

Thus,

Probability that the sum is 3 or less than 3 = 2/21

D) From the initial 21 outcomes, the number of outcomes that the sum is 2 or 10 are;

(1,1); (4,6) ; (5,5)

Thus,

Probability that the sum is 2 or 10 = 3/21 = 1/7

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