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Andrews [41]
2 years ago
5

Mr Kelly has a big pot of chili. He splits it into 9 bowls that each hold 1 1/8 cups. After filling those bowls, there is 1/2 cu

p left. How many cups did he start with?
Mathematics
1 answer:
nirvana33 [79]2 years ago
7 0
<h3>Answer:</h3>

10 5/8 cups

<h3>Explanation:</h3>

pot amount = 9 × (bowl amount) + (leftover amount)

... = 9 × (1 1/8) + 1/2

... = 9 × (1 + 1/8) + 4/8 . . . . . show 1 1/8 as a sum, rewrite 1/2 to 4/8

... = 9 + 9/8 + 4/8 . . . . . . . . eliminate parentheses using the distributive property

... = 9 + 13/8 . . . . . . . . . . . . add the fractions

... = 9 + 1 5/8 . . . . . . . . . . . convert from improper fraction to mixed number

... = 10 5/8 . . . . . cups . . . add the integer parts

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A set of face cards contains 4 Jacks, 4 Queens, and 4 Kings. Carlie chooses a card from the set, records the type of card, and t
iragen [17]

Answer:

Jan 5, 2017 - A set of face cards contains 4 Jacks, 4 Queens, and 4 Kings. Carlie chooses a card from the set, records the type of card, and then replaces the card. She repeats this procedure a total of 60 times. Her results are shown in the table. How does the experimental probability of choosing a Queen

Step-by-step explanation:

7 0
3 years ago
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Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children
Shkiper50 [21]

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error of the interval is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In this problem, we have that:

p = 0.69

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}

0.07\sqrt{n} = 1.2996

\sqrt{n} = \frac{1.2996}{0.07}

\sqrt{n} = 18.5658

(\sqrt{n})^{2} = (18.5658)^{2}

n = 345

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

4 0
2 years ago
A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p
Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so z = 1.645.

Item a:

The estimate is \pi = 0.213 - 0.195 = 0.018.

The sample size is <u>n for which M = 0.03</u>, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}

\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2

n = 53.1

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence \pi = 0.05

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 751.7

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

5 0
2 years ago
Please help..........
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