Ask question

Ask question

All categories

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?

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

You might be interested in

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

Read 2 more answers

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

3 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..........

wel

C = sin⁻¹ (sin(B) * c / b) = sin⁻¹ (sin [90degrees] * 6 / 12.00) = 30 degrees

3 0

2 years ago

HELP MEHHHHHHHHHHHHHHHHHHHHH xD

bezimeni [28]

I'm pretty sure the answer is D, if not C.

8 0

3 years ago

Read 2 more answers