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Arisa [49]
3 years ago
13

The gilberts are designing a rectangular patio for their backyard. the patio has an area of 432 square feet. the width of the pa

tio is ask your teacher its length. what is the length of the patio?
Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0
You didnt provide the width or the length.
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Which statement is true about the given information?
Archy [21]

Answer:

I think the answer is BD ≅ CE

8 0
3 years ago
The weights, in pounds, of a group of student are as follows: 173 123 171 175 188 120 177 160 151 169 162 128 145 140 158 132 20
Akimi4 [234]
Mean: 160.12
Standard Deviation: 19.76
Median: 162
Maximum: 202 
Minimum: 120

The mean is calculated by adding all of the numbers in the data set and dividing by the number of values that were added: 

173 + 123 + 171 + 175 + 188 + 120 + 177 + 160  + 151 + 169 + 162 + 128 + 145 + 140 + 158 + 132 + 202 + 162 + 154 + 180 + 164 + 166 + 157 + 171 + 175 ÷ 25 = 160.12

 Standard deviation is found by calculating the mean (160.12) , and then subtracting the mean and square root result for each number. Finally, working out the mean of the squared differences and taking the square root of the final answer results in a standard deviation of 19.76. (The attached image has the formula for finding standard deviation). 

Median is the number that is found in the middle when placing the numbers in order from least to greatest. 

The maximum is the largest number in the data set. 

The minimum is the smallest number in the data set.

5 0
4 years ago
Read 2 more answers
The living room floor measures 21 sq. yd.<br>How many square feet is that?<br>​
Alex_Xolod [135]

Answer:

21 square yard is 189 square feet.

7 0
3 years ago
Suppose the mean height of women age 20 years or older in a certain country is 62.7 inches. One hundred randomly selected women
zmey [24]

Answer: No, the data provided is not sufficient evidence to conclude that the mean height of women in the city differs from the national mean.

Step-by-step explanation:

Since we have given that

H_0:\mu =62.7\\\\H_a:\mu \neq 62.7

Mean height = 61.5 inches

Standard deviation = 4.4 inches

n = 100

So, test statistic value would be

z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{61.5-62.7}{\dfrac{4.4}{\sqrt{100}}}\\\\z=\dfrac{-1.2}{0.44}\\\\z=-2.727

At 10% level of significance, in two tail test ,

z = 1.28

Since 1.28 > -2.727

So, we will accept the null hypothesis.

Hence, No, the data provided is not sufficient evidence to conclude that the mean height of women in the city differs from the national mean.

7 0
4 years ago
A. 3:45<br> B. 4:00<br> C. 4:15<br> D. 4:45
aev [14]
Your  Answer is A:  3:45 
7 0
3 years ago
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