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2 years ago

Let x cm be the width of a rectangle whose length is 16 cm longer than its width.

Mathematics

1 answer:

nika2105 [10]2 years ago

6 0

We know that Length = x + 16 and Width = x.

Since the perimeter of the rectangle is 2 lengths + 2 widths,

Perimeter = 2(2x + 16) = 4x + 32.

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Jake tossed a paper cup 50 times and recorded how it landed. The table shows the results:

Dmitry [639]

Hello!

Here are 6 steps to Help You Out!

1. 3 divided by 50 = .06

2. Then if you like you can turn it in to a percent not sure how the answer is supposed to look for example:

3. 3/50=6/100=6%

4. 7 divided by 50 - .14

5. so 0.06 for the first one

6. And 40 divided by 50 = .8

====>So as we can see, open side up .06 closed side up .14 and on its side .8 or open 6% closed 14% and side 80% chance.<====

Hope this Helps! Have A WONDERFUL Day! :)

8 0

3 years ago

3. Find the product of 25x^2y and -6x²y3

mojhsa [17]

Answer:

-150 (x^2y+2)y³

Step-by-step explanation:

To Find the product of 25x^2y and -6x²y3.

(25x^2y) (-6x²y³)

= 25(-6) (x^2y+2) y³

= -150 (x^2y+2)y³

7 0

3 years ago

Six and four fifths divided by three and two ninths

ahrayia [7]

Its going to be 4 .............

4 0

3 years ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$