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kykrilka [37]
2 years ago
5

A group of students were surveyed about their interest in a new International Studies program. Interest was measured in terms of

high, medium, or low. 30 students responded high interest; 50 students responded medium interest; 20 students responded low interest. What is the relative frequency of students with high interest? A) 30% B) 50% C) 40% D) Cannot be de
Mathematics
1 answer:
Nady [450]2 years ago
7 0

Answer:

Relative frequency is the number of times an envent has occured.

When finding out the number of students with high interest you would take 30/100 because there was 30 students with high frequency and 100 students all together.

30/100=30%

A)30%

Hope this helps ;)

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Myra saved $15 a month for 18 months. She bought a book for $46.50 and a tennis racquet for $129.95. How much does she have left
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Answer:

A

Step-by-step explanation:

15*18=270.

270-46.50=223.5

223.5-129.95=93.55.

A. $93.55

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The green shaded region does not represent the solution to the system of linear inequalities given below. Explain why not.
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Answer:


Step-by-step explanation:

The area of solutions of the first inequality lies above an orange straight line.

The area of solutions of the second inequality lies below a blue straight line.

These areas are crossed in the yellow field.

The green field does not meet a condition of the second inequality and therefore does not enter the area of the solution of a system of inequalities

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Jessica shoebox is 20 cm long and 10 cm wide. How many more millimetres is the length of the shoebox than the width?
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8 0
2 years ago
According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m
tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

P(A) = 0.8x + 0.25*0.2x = 0.85x

Probability of developing diabetes while being very active:

0.25x of 0.2. So

P(A \cap B) = 0.25x*0.2 = 0.05x

What is the probability that a middle-aged man with diabetes is very active?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0
2 years ago
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