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Brums [2.3K]
3 years ago
8

Please help!! I can’t figure this out

Mathematics
1 answer:
Rudiy273 years ago
5 0
Each room has 2 walls that are 4x10 = 40 square feet and 2 walls that are 8x10 = 80 square feet. So 80+80+40+40 = 240 square feet of wall will be painted per room.

We use 2 coats of paint so in effect, we double the area. Imagine laying one thin wall on top of the other. The amount of paint area used per room doubles from 240 to 480

There are 4 of these rooms so 4*480 = 1920

The total area we need to paint is 1920 square feet

Divide this over 32 to get 1920/32 = 60

Final Answer: 60
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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be
Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

A = a + (A \cap B)

In which a is the probability that a student is proficient in reading but not mathematics and A \cap B is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

B = b + (A \cap B)

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

(A \cup B) + C = 1

In which

(A \cup B) = a + b + (A \cap B)

65% were found to be proficient in both reading and mathematics.

This means that A \cap B = 0.65

78% were found to be proficient in mathematics

This means that B = 0.78

B = b + (A \cap B)

0.78 = b + 0.65

b = 0.13

85% of the students were found to be proficient in reading

This means that A = 0.85

A = a + (A \cap B)

0.85 = a + 0.65

a = 0.20

Proficient in at least one:

(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98

What is the probability that the student is proficient in neither reading nor mathematics?

(A \cup B) + C = 1

C = 1 - (A \cup B) = 1 - 0.98 = 0.02

There is a 2% probability that the student is proficient in neither reading nor mathematics.

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3 years ago
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erastova [34]

Answer:

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add the stuff together

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