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wel
2 years ago
9

A die is thrown 100 times and a ‘6’ occurs on 32 occasions. Based upon this information find the probability of getting a ‘6’ wh

en the die is next rolled.
Mathematics
1 answer:
Morgarella [4.7K]2 years ago
3 0

in every die thrown 100 times a six occurs 32 times

so the probability of obtaining a die after 100 throws =  \frac{32}{100}


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Put in slope-intercept form would be: y<(3/4)x+1
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-4(2x + 12) in distributive property​
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−8x − 48

Step-by-step explanation:

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Construct a 95% confidence interval for a population proportion using repeated tests of significance to develop an interval of p
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2 years ago
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
2 years ago
g People between the ages of 65 and 70 were studied to see if those who exercise more were healthier. One measure of health was
liubo4ka [24]

Answer:

If the person selected had high blood pressure, the probability that he or she did not exercise is P=0.643.

Step-by-step explanation:

We have a group of people between agss of 65 and 70.

They are divided in two groups regarding their amount of exercise: "none", "moderate" and "high".

The response variable is blood pressure, that was categorized as "normal"or "high".

Then,  we have the frequencies table as:

Amount of exercise          None    Moderate    High

Blood pressure High          54              23            7

Blood pressure Normal      86              76            32

We have to calculate the probability that, given that the person had high blood pressure, he or she did not exercise.

If the person had high blood pressure, it can be of the any of the 3 groups of exercise amount. There are 54 people with high blood pressure that did not exercise, 23 that did moderate amount of exercise, and 7 that did a high amount.

Then, the probability can be calculated dividing the amount of people that had high blood pressure and did not exercise, by all the people that have high blood pressure (regarding the amount of exercise):

P(N | HP)=\dfrac{54}{54+23+7}=\dfrac{54}{84}=0.643

If the person selected had high blood pressure, the probability that he or she did not exercise is P=0.643.

5 0
3 years ago
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