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2 years ago

Add 1/10 + 2/15 (show steps pls-) :)

Mathematics

1 answer:

Ksivusya [100]2 years ago

3 0

Answer:

$\frac{7}{30}$

OR...

0.233333333

Step-by-step explanation:

Least common multiple of 10 and 15 is 30. Convert $\frac{1}{10}$ and $\frac{2}{15}$ to fractions with denominator 30.

$\frac{3}{30} +\frac{4}{30}$

Since $\frac{3}{30}$ and $\frac{4}{30}$ have the same denominator, add them by adding their numerators.

$\frac{3+4}{30}$

Add 3 and 4 to get 7.

$\frac{7}{30}$

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At a liberal arts college, 90% of the freshmen are enrolled in English 105, 80% are enrolled in Mathematics 101 and 5% are enrol

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a) There is a 75% probability that the freshman is enrolled in both English 105 and Mathematics 101.

b) There is a 15% probability that the freshman is enrolled in English 105 but not in Mathematics 101.

c) 93.75% probability that the freshman is also enrolled in English 105

Step-by-step explanation:

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

I am going to say that:

A are those freshmen enrolled in English 105.

B are those freshman Enrolled in Mathematics 101.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a freshmen is enrolled in English 105 but not in Mathematics 101 and $A \cap B$ is the probability that a freshmen is enrolled in both these classes.

By the same logic, we have that:

$B = b + (A \cap B)$

Solution

90% of the freshmen are enrolled in English 105

So A = 0.9

80% are enrolled in Mathematics 101

So B = 0.8

5% are enrolled in Mathematics 101 but not in English 105.

So b = 0.05

a. What is the probability that the freshman is enrolled in both English 105 and Mathematics 101?

This is $A \cap B$ .

$B = b + (A \cap B)$

$0.8 = 0.05 + (A \cap B)$

$A \cap B = 0.75$

There is a 75% probability that the freshman is enrolled in both English 105 and Mathematics 101.

b. What is the probability that the freshman is enrolled in English 105 but not in Mathematics 101?

This is a.

$A = a + (A \cap B)$

$0.90 = a + 0.75$

$a = 0.15$

There is a 15% probability that the freshman is enrolled in English 105 but not in Mathematics 101.

c. Suppose the freshman chosen is known to be enrolled in Mathematics 101. What is the probability that the freshman is also enrolled in English 105?

This is $P(A \cap B)$ divided by P(B). So 0.75/0.8 = 0.9375.

There is a 93.75% probability that the freshman is also enrolled in English 105

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