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user100 [1]
3 years ago
14

With a height of 65 ​in, William was the shortest president of a particular club in the past century. The club presidents of the

past century have a mean height of 71.6 in and a standard deviation of 1.1 in. a. What is the positive difference between William​'s height and the​ mean?
Mathematics
1 answer:
shusha [124]3 years ago
5 0

Answer:

according to my calculations he will be 69.6942069420 inches

Step-by-step explanation:

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Answer:

5,670

Step-by-step explanation:

how do you solve 135x42? (step by step please)

Answer :

= 135 x 42

= 5,670

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When solving -1/8(x+35)=-7 , what is the correct sequence of operations
TiliK225 [7]

x= -623/18

Hope it helps for you,buddy!


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100 POINTS!!!!!!!!!!!!!!!!!!!!!!!!
hodyreva [135]
<h3>Key points :-</h3>

✪ Both triangles will be proven similar by AA theorem i.e. Angle-Angle theorem.

✪ The symbol for similarity is ~.

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6 0
2 years ago
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The mean of the heights of three buildings A,B and C is 14.2m. The mean of the heights of five buildings A,B,C, D and E is 14.8m
Jobisdone [24]
<h3>Answer:  15.7 meters</h3>

========================================================

Work Shown:

s = sum of the heights of buildings A,B,C

s/3 = mean of the heights of buildings A,B,C

s/3 = 14.2

s = 3*14.2

s = 42.6

The buildings A,B,C have their heights add up to 42.6

--------------

r = sum of the heights of buildings A,B,C,D,E

r/5 = mean of the heights of buildings A,B,C,D,E

r/5 = 14.8

r = 5*14.8

r = 74

--------------

r-s = (sum of heights of A,B,C,D,E)-(sum of heights of A,B,C)

r-s = sum of the heights of buildings D and E

r-s = 74-42.6

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(r-s)/2 = mean height of buildings D and E

(r-s)/2 = 31.4/2

(r-s)/2 = 15.7 meters is the average height of buildings D and E.

4 0
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Nataly [62]

Answer:

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width: 8 cubes

length: 10 cubes

height: 7 cubes

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divided l, w, and h by 1/3 to get them in cubes

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