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maks197457 [2]
2 years ago
10

HELP ME PLEAS EI NEED THIS IMMEDAITEL

Mathematics
2 answers:
ella [17]2 years ago
5 0
Hope u did well bro?
Paladinen [302]2 years ago
3 0
Where is the question though
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Please help me???? I need ASAP
Assoli18 [71]

Answer:

D

Step-by-step explanation:

4 0
3 years ago
Two bowlers bowl the following number of strikes in 9 games
ehidna [41]
First, let's calculate the mean and the mean absolute deviation of the first bowler.

FIRST BOWLER: <span>8,5,5,6,8,7,4,7,6
Mean = (Sum of all data)/(Number of data points) = (8+5+5+6+8+7+4+7+6)/9
<em>Mean = 6.222</em>
Mean absolute deviation or MAD = [</span>∑(|Data Point - Mean|]/Number of Data Points
MAD = [|8 - 6.222| + |5 - 6.222| + |5 - 6.222| + |6 - 6.222| + |8 - 6.222| + |7 - 6.222| + |4 - 6.222| + |7 - 6.222| + |6 - 6.222|]/9
<em>MAD = 1.136</em>

SECOND BOWLER: <span>10,6,8,8,5,5,6,8,9
</span>Mean = (Sum of all data)/(Number of data points) = (<span>10+6+8+8+5+5+6+8+9</span>)/9
<em>Mean = 7.222</em>
Mean absolute deviation or MAD = [∑(|Data Point - Mean|]/Number of Data Points
MAD = [|10 - 7.222| + |6 - 7.222| + |8 - 7.222| + |8 - 7.222| + |5 - 7.222| + |5 - 7.222| + |6 - 7.222| + |8 - 7.222| + |9 - 7.222|]/9
<em>MAD = 1.531
</em>

The mean absolute deviation represents the average distance of each data to the mean. Thus, the lesser the value of the MAD is, the more consistent is the data to the mean. <em>B</em><em>etween the two, the first bowler is more consistent.</em>

8 0
3 years ago
Write a unit rate for: 12 pounds for $17.88.
Vika [28.1K]

Answer:

.67

Step-by-step explanation:

12/17.88=.67

8 0
3 years ago
Somebody please help me
Mazyrski [523]
Plan A: 30 + 0.01m
Plan B: 20 + 0.05m
in order for 2 plans cost the same, both need to be equal
so should be

20 + 0.05m = 30 + 0.01m 

answer is D. last one
20 + 0.05m = 30 + 0.01m 
8 0
2 years ago
A Web music store offers two versions of a popular song. The size of the standard version is 2.4 megabytes (MB). The size of the
ANEK [815]

Answer:

50.mb

Step-by-step explanation:

7 0
3 years ago
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