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Nimfa-mama [501]
3 years ago
10

The length of a rectangle is five times its width.

Mathematics
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{125 \:  {cm}^{2} }}}}}

Step-by-step explanation:

Let the width be 'w'

Length of a rectangle = 5w

Perimeter of a rectangle = 60 cm

Area of a rectangle = ?

<u>First</u><u>,</u><u> </u><u>finding </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>width </u><u>'</u><u> </u><u>w</u><u> </u><u>'</u>

\boxed{ \sf{perimeter \: of \: a \: rectangle = 2(l + w)}}

plug the values

⇒\sf{60 = 2(5w + w)}

Distribute 2 through the parentheses

⇒\sf{60 = 10w + 2w}

Collect like terms

⇒\sf{60 = 12w}

Swap the sides of the equation

⇒\sf{12w = 60}

Divide both sides of the equation by 12

⇒\sf{ \frac{12w}{12}  =  \frac{60}{12} }

Calculate

⇒\sf{w = 5} cm

<u>Finding </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>length</u><u> </u><u>(</u><u> </u><u>l</u><u> </u><u>)</u>

\sf{length = 5w}

Substitute the value of w

⇒\sf{length = 5 \times 5}

⇒\sf{length = 25 \: cm}

<u>Finally</u><u>,</u><u> </u><u>Finding</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> </u><u>rectangle</u><u> </u><u>having</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>2</u><u>5</u><u> </u><u>cm </u><u>and</u><u> </u><u>width </u><u>of</u><u> </u><u>5</u><u> </u><u>cm</u>

\boxed{ \sf{area \: of \:  a \: rectangle = l \times b}}

plug the values

⇒\sf{area \:  =  \: 25 \times 5}

Multiply the numbers

⇒\sf{area \:  =  \: 125 \:  {cm}^{2} }

Hope I helped!

Best regards!!

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Step-by-step explanation:

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Answer:

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Alternative hypothesis:\mu > 120  

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The p value for this case taking in count the alternative hypothesis would be:

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Step-by-step explanation:

Information given

\bar X=130 represent the sample mean for the amount spent each shopper

s=40 represent the sample standard deviation

n=80 sample size  

\mu_o =120 represent the value to verify

t would represent the statistic    

p_v represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

The statistic for this case would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Replacing the info given we got:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

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