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andre [41]
3 years ago
12

the perimeter of a rectangle is 66 cm and its width is half its length. what are the length and the width of the rectangle?

Mathematics
2 answers:
Sauron [17]3 years ago
8 0
P = 2(l + w)

w = l/2

Plug in 'l/2' for 'w' in the equation with '66':

66 = 2(l + l/2)

66 = 2l + 2l/2

66 = 2l + l

66 = 3l

22 = l

So the length is 22.

Now plug this in to find the width.

66 = 2(22 + w)

66 = 44 + 2w

22 = 2w

11 = w

So the width is 11, and the length is 22.
jok3333 [9.3K]3 years ago
8 0
Let's set width = w

Then length = 2w

We can set up the equation:

2(w+ 2w) = 66

2(3w) = 66

6w = 66

w = \frac{66}{6} = \boxed{11}

Then we plug in:

l = 2w = 2(11) = \boxed{22}

Hope this helps :)

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Read 2 more answers
A student records the repair cost for 22 randomly selected dryers. A sample mean of $98.78 and standard deviation of $15.49 are
Gelneren [198K]

Answer:

The critical value that should be used is T = 2.0796.

The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 22 - 1 = 21

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.0796, which is the critical value that should be used.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.0796\frac{15.49}{\sqrt{22}} = 6.868

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.78 - 6.868 = $91.912

The upper end of the interval is the sample mean added to M. So it is 98.78 + 6.868 = $105.648

The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.

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