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Yuki888 [10]
3 years ago
9

A square swimming pool has a cement sidewalk around it. The sidewalk is the same width all the way around. The outside perimeter

of the sidewalk is 80 feet. What is the width of the sidewalk if the area of the pool is 225 square feet?
Mathematics
1 answer:
forsale [732]3 years ago
7 0
I’m pretty sure u just have to subtract 225-80
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Answer:

m=1

Step-by-step explanation:

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10x - (5+3x)<br> To short LOL
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Harry has three vehicles (bicycle,scooter, and car). He uses them to go to school, the garden, and the theatre. If he is out of
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3 years ago
You would like to know whether forwards in a basketball league average the same (or different) numbers of points than the overal
Akimi4 [234]

Answer:

The critical t-value for such a test given an alpha level of 0.05 is 2.26

Step-by-step explanation:

Null hypothesis : H_0:\mu = 7.5

Alternate hypothesis :H_a:\mu \neq 7.5

Population mean = \mu = 7.5

Data : 10, 9, 6, 11, 13, 14, 9, 9, 9, and 10

Mean = \bar{x}=\frac{\text{Sum of all observations}}{\text{no. of observations}}

Mean =\frac{10+9+6+11+13+14+9+9+ 9+ 10}{10}

Mean =10

Standard deviation : \sqrt{\frac{\sum(x-\bar{x})^2}{n}}

Standard deviation :\sqrt{\frac{(10-10)^2+(9-10)^2+(6-10)^2+(11-10)^2+(13-10)^2+(14-10)^2+(9-10)^2+(9-10)^2+(9-10)^2+(10-10)^2}{10}}

Standard deviation s :2.144

t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}t = \frac{10-7.5}{\frac{2.144}{\sqrt{10}}}t=3.687

Df = n-1 = 10-1 = 9

T critical =t_{(df,\alpha)}=t_(9,0.05)=2.26

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So, We failed to accept null hypothesis

Hence the critical t-value for such a test given an alpha level of 0.05 is 2.26

5 0
3 years ago
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Step-by-step explanation:

The variation is direct, i.e. more the time of work, more the amount.

For 10 hours, the paycheck is $57.50,

so for 1 hour, the paycheck is $57.50 ÷ 10, or $5.75.

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p = 5.75 × t

Please mark Brainliest if this helps!

8 0
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