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Ber [7]
4 years ago
6

for the given perimeter find the length and width of the rectangle with the greatest area use whole numbers only

Mathematics
1 answer:
Viefleur [7K]4 years ago
8 0
So, the formula for the perimeter of a rectangle is 2l+2w=p. We know that p=80. Our options for w and l are: 30 and 10, 20 and 20, 25 and 15. To find the are we multiply the length by the width. 30*10=300, 20*20=400, 25*15=375. The greatest area is with the length and width both being 20.

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Tommy and Suzi spend much of the day sleeping. Tommy sleeps for 3/5 of the day and Suzi sleeps for 7/10 of the day.
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Answer:

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Step-by-step explanation:

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bulgar [2K]

Answer:

Step-by-step explanation:

Hello!

Out of 846 patients taking a prescription drug daily, 18 complained of flulike symptoms.

It is known that the population proportion of patients that take the drug of the competition and complain of flulike is 1.8%

Be the variable of interest:

X: number of patients that complained of flulike symptoms after taking the prescription drug, out of 846.

sample proportion p'= 18/846= 0.02

You have to test if the population proportion of patients that experienced flulike symptoms as a side effect is greater than 1.8% (p>0.018)

Assuming that the patients for the clinical trial were randomly selected.

The expected value for this  sample is np=846*0.02= 1658 (the expected value of successes is greater than 10) and the sample is less than 10% of the population, you can apply the test for the proportion:

The hypotheses are:

H₀: p ≤ 0.018

H₁: p > 0.018

α: 0.01

Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }≈N(0;1)

Z_{H_0}= \frac{0.02-0.018}{\sqrt{\frac{0.018*0.982}{846} } }= 0.437

The p-value for this test is  0.331056

The decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value is greater than α, the decision is to reject the null hypothesis.

So at 1% significance level there is no significant evidence to reject the null hypothesis, you can conclude that the population proportion of patients that took the prescription drug daily and experienced flulike symptoms as a side effect is less or equal to 1.8%

I hope this helps!

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