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Assoli18 [71]
3 years ago
11

QUICKLY HELP!Drag and drop the constant of proportionality into the box to match the table. If the table is not proportional, dr

ag and drop "not proportional" into the box.
x 2 4 6 8
y 3 7 11 15

possible answers:
1/3
not proportional
3/2
1
Mathematics
2 answers:
Tema [17]3 years ago
8 0
It is not proportional. (:
Ksivusya [100]3 years ago
8 0

Answer:

The asnwer is Not Proportional

Step-by-step explanation:

This is how:  You must divide Y by X in order to find answer.  All of the asnwers do come out differently though.

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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
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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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