The length of a diagonal of a square is 18√2 inches. Find the perimeter of the square.

1 answer:

3 0

A diagonal of a square is defined, $d=a\sqrt{2}$ where a is side of the square. Solve for a to get $a=\frac{d}{\sqrt{2}}$ .

Now multiply a by 4 since that is the definition of a perimeter,

$P=4a=4\cdot\frac{18\sqrt{2}}{\sqrt{2}}=4\cdot18=\boxed{72}$ .

Hope this helps.

You might be interested in

The arena wants to estimate the previous two month’s ticket-sale income. in total, 71,115 people attended different events at th

OleMash [197]

There was a total of 71,115 people who attended the different events in the arena for the past two months,

The estimated average ticket price is $ 21.15.

We need to solve for the estimated arena's ticket sale income and the solution is shown below:
Ticket Sale Income = 71, 115 * $21.15
Ticket Sale Income = $1,504,082

3 0

3 years ago

I need help As Soon As possible!

Julli [10]

This should be simple, add the Amounts of snow for the first week
4.32 + 6.86 = 11.18
-------------------------------------------------------
The subtract the amounts of snow from each week
11.18 - 7.89 = 3.29
----------------------------------------
C.) 3.29 is your answer

6 0

3 years ago

A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before t

aleksklad [387]

Answer:

We cannot say that the mean wake time are different before and after the treatment, with 98% certainty. So the zopiclone doesn't appear to be effective.

Step-by-step explanation:

The goal of this analysis is to determine if the mean wake time before the treatment is statistically significant. The question informed us the mean wake time before and after the treatment, the number of subjects and the standard deviation of the sample after treatment. So using the formula, we can calculate the confidence interval as following:

$IC[\mu ; 98\%] = \overline{y} \pm t_{0.99,n-1}\sqrt{\frac{Var(y)}{n}}$

Knowing that $t_{0.99,15} = 2.602$ :

$IC[\mu ; 98\%] = 98.9 \pm 2.602\frac{42.3}{4} \Rightarrow 98.9 \pm 27.516$

$IC[\mu ; 98\%] = [71.387 ; 126,416]$

Note that $102.8 \in [71.384 ; 126.416]$ so we cannot say, with 98% confidence, that the mean wake time before treatment is different than the mean wake time after treatment. So the zopiclone doesn't appear to be effective.