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Crazy boy [7]
2 years ago
15

Pleaseeeee helppp asap

Mathematics
2 answers:
Ber [7]2 years ago
5 0

Answer:

35.7 megabytes

Step-by-step explanation:

Let's first find 10% of the percent needed. If it is 234 megabytes and 10% has downloaded so far, we conclude that 10% of the download is 23.4 megabytes (since we move the decimal point one spot to the left to find ten percent).

To get to 15% of the percent needed, let's split 10% in half to get the amount of megabytes per every 5%.

23.4/2 = 11.7

We can now conclude that 5% = 11.7. Therefore, 15% will equal (23.4 + 11.7) 35.1 megabytes.

To find the remaining 3.5% left of the percent needed, we use a calculator.

3.5% x 18 = 0.63.

We will add that to the 35.1 megabytes, but we need to round to the nearest tenth, as the question stated. Therefore, we round to 0.6, add it to 35.1, to get a total of 35.7 megabytes.

For 18.5% of the download, it will be 35.7 megabytes.

wlad13 [49]2 years ago
3 0

Answer:

Step-by-step explanation:

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Y is a random variable that is distributed N(-16, 1.21). Find k such that Prob(-15.043 < Y ≤ k) = 0.1546. (Round your answer
IgorLugansk [536]

Transform <em>Y</em> to <em>Z</em>, which is distributed N(0, 1), using the formula

<em>Y</em> = <em>µ</em> + <em>σZ</em>

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Pr[-15.043 < <em>Y</em> ≤ <em>k</em>] = 0.1546

Pr[(-15.043 + 16)/1.21 < (<em>Y</em> + 16)/1.21 ≤ (<em>k</em> + 16)/1.21] = 0.1546

Pr[0.791 < <em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.1546

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] - Pr[<em>Z</em> < 0.791] = 0.1546

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] = 0.1546 + Pr[<em>Z</em> < 0.791]

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.1546 + 0.786

Pr[<em>Z</em> ≤ (<em>k</em> + 16)/1.21] ≈ 0.940

Take the inverse CDF of both sides (<em>Φ(x)</em> denotes the CDF itself):

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