Lee needed to make drapes the length from the floor to the top of the window. The wall beneath the window had the height of 2 2/

3 feet and the height of just the window measured 2 3/4 feet. How long do the drapes have to be?

Mathematics

1 answer:

aalyn [17]2 years ago

3 0

$We\ should\ sum \ those \ two\ measurements\ to \ obtain\ length\ of\ drapes\\\\
2 \frac{2}{3}+2\frac{3}{4}=\frac{8}{3}+\frac{11}{4}=\frac{24}{12}+\frac{33}{12}=\frac{57}{12} =4,75feet$

You might be interested in

Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?

satela [25.4K]

Answer:

$755.80

Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

The compound amount formula is A = P (1 + r/n)^(nt), where

P is the initial principal, r is the interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years. Here, P = $2179; t = 5 yrs; r = 0.06; and n = 4 (quarterly compounding).

We get:

A = $2179(1 + 0.06/4)^(4*5), or $2179(1.015)^20, or $2179(1.347) = $2937.80.

The compound amount is $2934.80. Subtracting the $2179 principal results in the interest earned: $755.80.

5 0

2 years ago

Throughout the 2016 Presidential election primaries, Millennials (those aged 20 t 36 years) consistently supported Senator Berni

Dmitrij [34]

Answer:

The 90% confidence interval for the support of Sanders by millenials is between 53% and 57%.

$0.53 \leq p \leq 0.57$

Step-by-step explanation:

In this question we have to calculate a confidence interval (90% CI) on the proportion of millenials that had a favorable opinion on Sanders.

The sample size is n=1,754.

The p-hat, taken from the poll, is

$\hat{p}=0.55$

The estimated standard deviation is equal to:

$\sigma_p=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.55*0.45}{1754}}=0.012$

For a 90% CI, the z-value is z=1.645.

Then, the confidence interval is

$\hat{p}-z*\sigma_p\leq p\leq\hat{p}-z*\sigma_p\\\\0.55-1.645*0.012\leq p\leq 0.55-1.645*0.012\\\\0.55-0.02 \leq p \leq 0.55+0.02\\\\0.53 \leq p \leq 0.57$