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jonny [76]
2 years ago
7

Solve for x

Mathematics
1 answer:
Ghella [55]2 years ago
7 0

Answer:

1. x=105

2. 12 cans for $4

3. 14

Step-by-step explanation:

1. 15/18 = x/126

126/18=7

x=15*7=105

2. 12 cans for $4.  36 for $11.50.  

36/3=12.  11.50/3=3.8333.  

12 cans for $4 is the better deal because it would be an equal deal if it was 36 for 12:00.  

3. Circumference = pi*d.  Circumference = 43.96.  pi=3.14.  3.14*x=43.96.  x=diameter.  x=14.  diameter = 14

--Explanation to 3.14*x=43.96:

3.14*x=43.96, = 3.14x=43.96

3.14x*100 =43.96*100

314x=4396

314x/314 = 4396/314

x=14

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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

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(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

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

This means that A = 0.85

A = a + (A \cap B)

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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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