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3 years ago

You buy two packages of almonds. One package is 10.47 kg and the other is 8.23 kg. What is the difference?

Mathematics

1 answer:

fomenos3 years ago

5 0

Answer:

2.04 kg

Step-by-step explanation:

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Grade 5 module 2 lesson 25 problem set

Marizza181 [45]

I dont see a question so i can’t answer

5 0

3 years ago

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the

Vesna [10]

Answer:

A sample size of at least 1,353,733 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , 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 .

The margin of error is:

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

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?

We need a sample size of at least n.

n is found when M = 0.001.

Since we don't have an estimate for the proportion, we use the worst case scenario, that is $\pi = 0.5$

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

$0.001 = 2.327\sqrt{\frac{0.5*0.5}{n}}$

$0.001\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.001}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.001})^{2}$

$n = 1353732.25$

Rounding up

A sample size of at least 1,353,733 is required.

5 0

3 years ago

Elijah is selling coupon books for a school fundraiser. the coupon books sell for $11. the school gets $8 and $3 goes to elijah'

Sladkaya [172]

He collected $264.00, $72.00 went to his homeroom, and $192.00 to the school

6 0

3 years ago

In a sample of 53 walking canes, the average length was found to be 34.3in. with a variance of 1.9. Give a point estimate for th

Anna11 [10]

Answer:

1.38

Step-by-step explanation:

The point estimate for the population standard deviation of the length of the walking canes is given as the squareroot of variance. Since variance is provided as 1.9 then the standard deviation will be

$\sqrt 1.9=1.37840487520902\approx 1.38$