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Mashcka [7]
3 years ago
10

X+3<8 and 3(x+4)-11<10

Mathematics
1 answer:
kondaur [170]3 years ago
4 0

Answer:

<h3>X<5 and X<3</h3>

Step-by-step explanation:

To solve this problem, first, you have to isolate it on one side of the equation. Remember to solve this problem, isolate x on one side of the equation.

x+3<8 and 3(x+4)-11<10

x+3<8

x+3-3<8-3 (First, subtract 3 from both sides.)

8-3 (Solve.)

8-3=5

x<5

3(x+4)-11<10

3(x+4)-11+11<10+11 (Add 11 from both sides.)

10+11 (Solve.)

10+11=21

3(x+4)<21

3(x+4)/3<21/3 (Next, divide by 3 from both sides.)

21/3 (Solve.)

21/3=7

x+4<7

x+4-4<7-4 (Then, subtract 4 from both sides.)

7-4=3

x<3

The correct answer is x<5 and x<3.

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Answer:

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Step-by-step explanation:

Given:

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But the question mentions TWO sites, hence there will be two site visits. Therefore for TWO sites, we have to account for TWO site visit fee.

Therefore for TWO sites, the total amount of money earned

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We are further given that we worked y hours at the hourly work rate of $24 per hour.

Hence the cost for working

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= $24y   (substituting this into equation 1 above)

for TWO sites, the total amount of money earned

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Answer:

The number of students that like only two of the activities are 34

Step-by-step explanation:

Number of students that enjoy video games, A = 38

Number of students that enjoy going to the movies, B = 12

Number of students that enjoy solving mathematical problems, C = 24

A∩B∩C = 8

Here we have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) -n(A∩C) + n(A∩B∩C)

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Also the number of student that like only one activity is found from the following equation;

n(A) - n(A∩B) - n(A∩C) + n(A∩B∩C) + n(B) - n(A∩B) - n(B∩C) + n(A∩B∩C) + n(C) - n(C∩B) - n(A∩C) + n(A∩B∩C) = 30

n(A) + n(B) + n(C) - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 3·n(A∩B∩C) = 30

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- 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) = -68

n(A∩B) + n(B∩C) + n(A∩C) = 34

Therefore, the number of students that like only two of the activities = 34.

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