The perimeter of a rectangle is 142m. The length is 2m more than the width. Find the dimensions. What is the length?

1 answer:

5 0

You have a rectangle that has a width of w. the length us 2 more than the width, so the lengthis w+2. the perimeter is 142. so 142=w+w+(w+2)+(w+2). because the perimeter is all 4 sides added together 2 of those sides being the length and 2 beingtje width. combine like terms to get 142=4w+4. now you subtract 4 from each side and get 138=4w. divide both sides by 4. you get w=34.5. now, w is the width. the length is 2 more than w, which is 34.5+2, or 36.5

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5−4+7+1= 5 - 4 + 7 x + 1 = how is it no solution

laila [671]

Answer:

There is a solution. Here's how:

Step-by-step explanation:

So, you work out the problem, and you lay it out here:

5-4+7+1=5-4+7x+1.

First, you start by subtracting 5 from both sides, which cancels both of the 5s. Here's your equation now:

-4+7+1=-4+7x+1

Next, you add 4 to both sides, which makes you identify the identity.

Here's your equation now:

7+1=7x+1

Now, you subtract 1 from both sides.

Here's your equation now:

7=7x

Finally, divide both sides by 7, while identifying the identity.

x=1.

8 0

3 years ago

20 pts to whoever helps!!

VladimirAG [237]

<h3>Answer: B) 31.348</h3>

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Work Shown:

We will never use the alpha = 0.05 value when computing the test statistic. So we can ignore alpha for this particular problem.

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n = 40 is the sample size.

s = 2.6 is the sample standard deviation

$\sigma$ = sigma = 2.9 is the population standard deviation (ie the claimed standard deviation for all the pay phones)

Test Statistic for chi-square

$\chi^2 = \frac{(n-1)*s^2}{\sigma^2}\\\\\chi^2 = \frac{(40-1)*(2.6)^2}{(2.9)^2}\\\\\chi^2 = \frac{39*6.76}{8.41}\\\\\chi^2 = \frac{263.64}{8.41}\\\\\chi^2 \approx 31.3483947681332\\\\\chi^2 \approx 31.348\\\\$