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Lelechka [254]
3 years ago
7

Compute the average (mean) from the data shown. (round to the nearest tenth)

Mathematics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

C

Step-by-step explanation:

The mean is the average so you must add them all up and divide them by the amount of numbers there are.

Alexus [3.1K]3 years ago
6 0

Answer:

(C) 75.1

Step-by-step explanation:

Add all the numbers then divide by numbers of numbers their are

91+91+91+97+65+37+41+76+ 87= 676

676/9= 75.1111111111 and it goes on forever

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Did I do this right? If not please explain!
andrew11 [14]
Yep! Good job, you did it correctly.
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5 0
2 years ago
The temperature drops 17° to 37° . what was the temperature when we started?
Alex Ar [27]
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3 0
3 years ago
Read 2 more answers
By what number should-15/56 be divided to get -5/6<br><br>pls pls pls pls I need your help ​
mr_godi [17]

Answer:

The desired divisor is n = 21

Step-by-step explanation:

Let that number be n.  Then:

-15/56        -5

---------- = ---------

     n            6

Through cross-multiplication we get:

(-15/56)*6 = -5n, which reduces to:

(-15)(7) = -5n, or 3(7) = n

The desired divisor is n = 21

6 0
3 years ago
Wins 1975, loses 339, ties 10. What is my percent of wins?
dimaraw [331]

1975/1975+339+10

Wins/Total

1975/2324=0.8498

Convert to percent by moving decimal point up 2

Percent of wins: 84.98%

6 0
3 years ago
Probability of numbers divisible by 9 from numbers 1-100?
quester [9]
11
(9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99)
Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100:

100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99)
→ There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9.

(They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)
7 0
3 years ago
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