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Mariana [72]
3 years ago
10

The length of a rectangle is 5 meters greater than the width. The perimeter is 150 meters. Find the width AND the length.

Mathematics
1 answer:
oee [108]3 years ago
7 0

Answer:

L=62.5 W=12.5

Step-by-step explanation:

L=5W

150=2W+2(5W)

75=W+5W

75=6W

W=12.5

12.5×5=62.5

62.5+62.5+12.5+12.5=150

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Hello,
Answer B

If x≠y then
\dfrac{x-y}{-x+y}=\dfrac{x-y}{-(x-y)}=-1


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Abargain clothing store is selling 5 shirts for $20.
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the length of a Ribbon is 11.55 metre if it is cut into pieces and divided equally among 7 Girls how much ribbon will each get f
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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
X is 13 less than 17. WRITE AN EQUATION​ TO REPRESENT THE statement.
vovikov84 [41]

Answer:

X=4

Step-by-step explanation:

13 less than 17 would be 4 becayse if you do 17-13=4

6 0
3 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.
garik1379 [7]

Answer:

The 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =9.2 represent the sample mean 1

\bar X_2 =8.8 represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

\sigma_1 =0.3 population standard deviation for sample 1

\sigma_2 =0.1 population standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =9.2-8.8=0.4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281  

0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519  

So on this case the 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

6 0
3 years ago
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