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vovikov84 [41]
3 years ago
11

Alex accidentally forgot to stock up on toilet paper before the stay-at-home order. Now he has to buy toilet paper on the black

market. Though the price of toilet paper on the black market has mostly stabilized, it still varies from day to day. The daily price of a generic brand 12-pack, X, and the daily price of a generic brand 6-pack, Y, (in rubles) jointly follow a bivariate normal distribution with:
μx = 2,470, σx = 30, μy = 1,250, σ = 25, p = 0.60.
(a) What is the probability that 2 (two) 6-packs cost more than 1 (one) 12-pack? (b) To ensure that he will not be without toilet paper ever again, Alex buys 7 (seven) 12-packs and 18 (eighteen) 6-packs. What is the probability that he paid more than 40,000 rubles?
(c) Suppose that today's price of a 12-pack is 2,460 rubles. What is the probability that a 6-pack costs less than 1,234 rubles today? [1 US dollar is approximately 75 rubles ]
Mathematics
1 answer:
gladu [14]3 years ago
5 0
<h2>Alex accidentally forgot to stock up on toilet paper before the stay-at-home order. Now he has to buy toilet paper on the black market. Though the price of toilet paper on the black market has mostly stabilized, it still varies from day to day. The daily price of a generic brand 12-pack, X, and the daily price of a generic brand 6-pack, Y, (in rubles) jointly follow a bivariate normal distribution with: </h2><h2>μx = 2,470, σx = 30, μy = 1,250, σ = 25, p = 0.60. </h2><h2>(a) What is the probability that 2 (two) 6-packs cost more than 1 (one) 12-pack? (b) To ensure that he will not be without toilet paper ever again, Alex buys 7 (seven) 12-packs and 18 (eighteen) 6-packs. What is the probability that he paid more than 40,000 rubles? </h2><h2>(c) Suppose that today's price of a 12-pack is 2,460 rubles. What is the probability that a 6-pack costs less than 1,234 rubles today? [1 US dollar is approximately 75 rubles ]</h2>
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hammer [34]

Answer:

B. x ≥ 3 or x ≤ −2

Step-by-step explanation:

<u>Inequalities</u>

Solve the inequality:

x^2-x\ge 6

Subtracting 6:

x^2-x-6\ge 0

Factoring:

(x-3)(x+2) ≥ 0

We have a product that must be greater or equal to 0. This can only happen if:

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Or:

x - 3 ≤ 0  and  x + 2 ≤ 0

The first couple of conditions yields to:

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Which lead to the solution

x ≥ 3       [1]

The second couple of conditions yields to:

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Which lead to the solution

x ≤ -2        [2]

The final solution is [1] or [2]:

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B. x ≥ 3 or x ≤ −2

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3 years ago
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hmm

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ah ha

so we just need to find a and b such that b>a


b=10
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|a-b|=|5-10|=|-5|=5

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tada


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