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poizon [28]
3 years ago
13

The following gives information about the proportion of a sample that agree with a certain statement. Use StatKey or other techn

ology to find a confidence interval at the given confidence level for the proportion of the population to agree, using percentiles from a bootstrap distribution. StatKey tip: Use ‘‘CI for Single Proportion" and then ‘‘Edit Data" to enter the sample information.Find a 90% confidence interval if 112 agree and 288 disagree in a random sample of 400 people.What is the 90% confidence interval? ________, ________
Mathematics
1 answer:
marishachu [46]3 years ago
3 0

Answer:

(0.2278, 0.3322)

Step-by-step explanation:

Given that out of 400 people 112 agree and 288 disagree

Proportion of people agreeing = \frac{112}{400} =0.28

q=1-p =0.72\\se = \sqrt{\frac{pq}{n} } =0.03175

For confidence interval 90% we have critical value as

1.645

Margin of error =1.645*SE

=1.645(0.03175)\\= 0.0522

Confidence interval = (0.28-0.0522, 0.28+0.0522)\\= (0.2278, 0.3322)

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1920÷12= 160

Step-by-step explanation:

1920÷12= 160

4 0
3 years ago
Please help me with the question below
nata0808 [166]

Answer:

4

Step-by-step explanation:

If Judge is x years old and Eden is 6 years older, then Eden is x + 6 years old.

The second part tells us that Eden will be twice as old as Judge in two years.

This means that in two years: (Eden's age) = 2 * (Judge's age).

Since we know that Eden's age can be represented as x + 6 and Judge's age can be represented as x, we can write this: x + 6 = 2 * x

Simplify the equation:

x + 6 = 2x

6 = x = Judge's age (in two years)

If Judge is 6 two years later, then he must be 4 now.

To check our work, we can just look at the problem. Judge is 4 years old and Eden is 6 years older than Judge (that means Eden is 10 right now). Two years later, Eden is 12 and Judge is 6, so Eden is twice as old as Judge. The answer is correct.

6 0
3 years ago
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Yakvenalex [24]

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3 0
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Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

3 0
3 years ago
Which graph best represents the line y= -1/5x+2
vodomira [7]

Answer:

B

Step-by-step explanation:

Line crosses y-axis at 2.

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For each 1 square the line rises/falls it moves to the right/left 5 squares.

Negative slope lines are downhill left to right.  

8 0
3 years ago
Read 2 more answers
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