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neonofarm [45]
4 years ago
6

Which variable expression represents the word phrase? the sum of 12 and the quotient of 9 and a number

Mathematics
2 answers:
iris [78.8K]4 years ago
6 0
"Sum" tells us it's an addition problem:

One term is 12 and the other is 9/x

SO

The expression is:
12+9/x
Evgesh-ka [11]4 years ago
3 0

Answer:

The required expression is 12+\frac{9}{y}.

Step-by-step explanation:

Consider the provided phrase.

The sum of 12 and the quotient of 9 and a number

Let the number is y and the quotient of 9 and a number can be written as:

\frac{9}{y}

Thus, the sum of 12 and the quotient of 9 and a number can be written as:

12+\frac{9}{y}

Hence, the required expression is 12+\frac{9}{y}.

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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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sashaice [31]

Answer:

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Step-by-step explanation:

50/12 = 4.1666667. Round up, so 4.1666667 becomes 4.2.

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Step-by-step answer:

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