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2 years ago

2) f(x) = 27x^6+ 1524x^3+125

Mathematics

1 answer:

Norma-Jean [14]2 years ago

8 0

Answer:

Step-by-step explanation:

Let x^3 = y

f(x) = 27(x^3)^2 + 1524(x^3) + 125

f(x) = 27 y^2 + 1524y + 125

You can now put this into the quadratic formula. When you do, you get

(y + 0.0821)(y+56.3623) = 0

So now you substitute x^3 into this

(x^3 + 0.0821)(x^3 + 56.3623)

x^3 + 0.0821 = 0

x^3 = - 0.0821

cuberoot x^3 = cube root 0.0821

x = - 0.4346

x^3 = - 56.3623

x = - 3.8341

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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

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

Xbar = 377999 / 36 = 10499.9722

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

Xbar = 395197 / 36 = 10977.6944

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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2 years ago

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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3 years ago