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WINSTONCH [101]
4 years ago
13

NASA is painting the nose cone of a sounding rocket with a special sealant which reduces the air-drag on the rocket. If they nee

d to do two coats of the sealant, how many square feet are they painting? Use π = 3.14.
Mathematics
1 answer:
crimeas [40]4 years ago
4 0

Answer:

The formula for the lateral surface area (LSA) of a right cone is:

LSA = π x r x l

where: r as radius, and l as the slant height of the cone

If NASA need to do two coats of the sealant, the number of square feet that they are painting is: 2 x LSA = 2 x π x r x l

Step-by-step explanation:

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Natasha2012 [34]

Answer:

Snail A: 0, 5, 10, 15, 20, 25, 30

Snail A travels at a rate of 30 inches in 1 hour or 30 inches per hour.

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Snail B travels at a rate of 18 inches in 1 hour or 18 inches per hour.

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Snail C travels at a rate of 36 inches in 1 hour or 36 inches per hour.

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3 years ago
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.

6 0
3 years ago
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