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Pavlova-9 [17]
3 years ago
15

Estimate each percent. 35% of 147=

Mathematics
2 answers:
QveST [7]3 years ago
8 0
35\% \ of \ 147= \frac{35}{100}*147= \frac{7}{20}*147= \\\\ =\boxed{\frac{1029}{20}=51.45}
inessss [21]3 years ago
4 0
35\%*147\\\\
Write\ 35\%\ as\ \frac{35}{100}=0,35\\\\
0,35*147=51,45\\\\35\%\ of\ 147\ is\ equal\ to\ 51,45.
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Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Website 1 : 10357, 10537, 10767, 10561, 10544, 10581, 10602, 10665, 10335, 10419, 10737, 10410, 10485, 10601, 10458, 10472, 10435, 10375, 10436, 10510, 10345, 10559, 10520, 10425, 10351, 10465, 10491, 10671, 10366, 10440, 10618, 10606, 10406, 10538, 10449, 10462

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Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 1 :), s = 110.239865

Website 2 : 11067, 11029, 10888, 10789, 10914, 10663, 10787, 11140, 11042, 11074, 10868, 10853, 10900, 11088, 10991, 10928, 10959, 11126, 11033, 11114, 11150, 11155, 11027, 10900, 11015, 11123, 10953, 11181, 10855, 10731, 10971, 10770, 11070, 11122, 11018, 10903

Mean, xbar = ΣX/ n ; n = sample size = 36

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Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 2), s = 132.617995

2.)

Yes, the viewership between the two websites are different with the second website has a higher mean viewership with a mean of 10977.6944.

3.)

The probability of 12000 views per month on each website :

Probability = Mean viewership per month / required viewership

Website 1 :

P(12000) = 10499.9722 / 12000 = 0.8749

Website 2 :

P(12000) = 10977.6944 / 12000 = 0.9148

4.)

More consistent website :

We use the standard deviation value, the higher the standard deviation, the higher the variability :

Website 1 should be more consistent has it has a Lower standard deviation score, hence, should show lower variability than website 2.

5.)

Website suitable for advertisement should be one with higher viewership per month in other to reach a larger audience. Hence, website 2 should be recommended for advertisement.

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