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lawyer [7]
2 years ago
10

A local university wanted to understand what students prefer to eat during finals. They asked 1,000 students, "Do you prefer chi

cken, burgers, or pizza?" The results of the survey are shown in the two-way table below:
Chicken Burgers Pizza
Male 127 218 125
Female 219 192 119
What is the probability that a person chosen at random from this survey prefers pizza, given that they are female? Round your answer to the nearest tenth. (2 points)



11.9%
12.5%
22.5%
48.8%
Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
4 0
This is an example of conditional probability, which is the probability of an event occurring GIVEN the occurrence of some other event. There is a formula for this P(A|B)= \frac{P(A and B)}{P(B)} 

You want to find the probability<span> that a person chosen at random from this survey prefers pizza, given that they are female. so P(A and B) would refer to the proability that a student is female and prefers pizza. Look at the table and find the intersection of these two events and you would find the probability is </span>\frac{119}{1000}. Now find P(B) which is the probability that a student is female. Go to the table, and add up all the boxes where you see a student is female and you would find that P(B) is equal to \frac{530}{1000}. Now you wan to plug in these fractions into the formula to find P(A|B) and you get \frac{119}{1000} /\frac{530}{1000} =  \frac{119}{530} = 0.2245.. = 22.5 %

kompoz [17]2 years ago
3 0
Given that the person is female, the universe is reduced to 219 + 192 + 119 = 530 people.

The number of women that prefer pizza is 119.

Then the probability is 119/530 *100 = 22.5%
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The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches. Use the Empiric
maks197457 [2]

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between \text{Mean}\pm\text{Standard deviation}

i.e. about 68% of the heights of students lies between 55\pm5

Here, 55\pm5=(55-5, 55+5)=(50,60)

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

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2 years ago
The area of Kamila’s rectangular living room is 2.5 times the area of her square bedroom. The length of the living room is 18 fe
Archy [21]
Let L be the length of one side of Kamila's bedroom (since her bedroom is square, the area would be L x L). Then the width of her living room is 1.25L. So:
18*1.25L=2.5(L)²
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L=22.5/2.5=9
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2 years ago
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If your front lawn is 17.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snowflakes every minu
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Answer: 42.84 kg

Step-by-step explanation: Lawn is 17 ft x 20 ft.

Its area is S = 17 x 20 = 340 ft²

Each ft² of lawn accumulates 1050 snowflakes/min, 1ft² = 1050 s.f./min.

As 1 hour has 60 min, the lawn accumulates 63000/hour →

1050 x 60 = 63000

1ft² = 63000s.f./hour

This way, 340 ft² of lawn will accumulate 21,420,000 s.f/hour →

340 x 63000 = 21420000

As 1 snowflake has 2mg,  21,420,000 will have 42,840,000 mg.

As 1kg = 1,000,000 mg

42,840,000/1,000,000 = 42.84 kg

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A man bought a mobile phone for $800 and sold it for $1000. What was his profit as a percentage of the cost price ​
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Answer:

1000/800 = 125% profit

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In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. A long-distance telephone c
scoundrel [369]

Answer:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

Step-by-step explanation:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

And in order to test the hypothesis we can use a one sample t test or z test depending if we know the population deviation or not

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