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guajiro [1.7K]
4 years ago
14

Twenty five heat lamps are connected in a greenhouse so, that when one lamp fails, another takes over immediately (only one lamp

is turned on at any time.) The lamps operate independently, each with mean lifetime of 50 hours, and standard deviation of 4 hours. If the greenhouse is not checked for 1300 hours after the lamp system is turned on, what is the probability, that a lamp will be burning at the end of the 1300-hour period?
Mathematics
1 answer:
olga55 [171]4 years ago
4 0

Answer:

P = 0.006

Step-by-step explanation:

Given

n = 25 Lamps

each with mean lifetime of 50 hours and standard deviation (SD) of 4 hours

Find probability that the lamp will be burning at end of 1300 hours period.

As we are not given that exact lamp, it means we have to find the probability where any of the lamp burning at the end of 1300 hours, So we have

Suppose i represents lamps

P (∑i from 1 to 25 (X_i > 1300)) = 1300

= P(X^{'}> \frac{1300}{25})                  where X^{'} represents mean time of a single lamp

= P (Z> \frac{52-50}{\frac{1}{\sqrt{25}}})                Z is the standard normal distribution which can be found by using the formula

Z = Mean Time (X^{'}) - Life time of each Lamp (50 hours)/ (SD/\sqrt{n})

Z = (52-50)/(4/\sqrt{25}) = 2.5

Now, P(Z>2.5) = 0.006 using the standard normal distribution table

Probability that a lamp will be burning at the end of 1300 hours period is 0.006

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Answer:

<em>x = 2</em>

Step-by-step explanation:

We are given the equation 3 - 4x - 2 = 2x - 11;

3 - 4x - 2 = 2x - 11,\\3 - 2 = 6x - 11,\\1 = 6x - 11,\\12 = 6x,\\\\x = 2\\Conclusion, x = 2

If you want this explanation in a " statement reasoning " format refer to the attachment below;

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3 years ago
Compute the number of ways to deal each of the following five-card hands in poker. 1. Straight: the values of the cards form a s
Elenna [48]

Answer:

The number of ways to deal each hand of poker is

1) 10200 possibilities

2) 5108 possibilities

3) 40 possibilities

4) 624 possibilities

5) 123552 possibilities

6) 732160 possibilities

7) 308880 possibilities

8) 267696 possibilities

Step-by-step explanation:

Straigth:

The Straight can start from 10 different positions: from an A, from a 2, 3, 4, 5, 6, 7, 8, 9 or from a 10 (if it starts from a 10, it ends in an A).

Given one starting position, we have 4 posibilities depending on the suit for each number, but we need to substract the 4 possible straights with the same suit. Hence, for each starting position there are 4⁵ - 4 possibilities. This means that we have 10 * (4⁵-4) = 10200 possibilities for a straight.

Flush:

We have 4 suits; each suit has 13 cards, so for each suit we have as many flushes as combinations of 5 cards from their group of 13. This is equivalent to the total number of ways to select 5 elements from a set of 13, in other words, the combinatorial number of 13 with 5 {13 \choose 5} .  However we need to remove any possible for a straight in a flush, thus, for each suit, we need to remove 10 possibilities (the 10 possible starting positions for a straight flush). Multiplying for the 4 suits this gives us

4 * ( {13 \choose 5} -10) = 4* 1277 = 5108

possibilities for a flush.

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We have 4 suits and 10 possible ways for each suit to start a straight flush. The suit and the starting position determines the straight flush (for example, the straight flush starting in 3 of hearts is 3 of hearts, 4 of hearts, 5 of hearts, 6 of hearts and 7 of hearts. This gives us 4*10 = 40 possibilities for a straight flush.

4 of a kind:

We can identify a 4 of a kind with the number/letter that is 4 times and the remaining card. We have 13 ways to pick the number/letter, and 52-4 = 48 possibilities for the remaining card. That gives us 48*13 = 624 possibilities for a 4 of a kind.

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We need to pick the pair of numbers that is repeated, so we are picking 2 numbers from 13 possible, in other words, {13 \choose 2} = 78 possibilities. For each number, we pick 2 suits, we have {4 \choose 2} = 6 possibilities to pick suits for each number. Last, we pick the remaining card, that can be anything but the 8 cards of those numbers. In short, we have 78*6*6*(52-8) = 123552 possibilities.  

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We choose the number that is matching from 13 possibilities, then we choose the 2 suits those numbers will have, from which we have 4 \choose 2 possibilities. Then we choose the 3 remaining numbers from the 12 that are left ( 12 \choose 3 = 220 ) , and for each of those numbers we pick 1 of the 4 suits available. As a result, we have

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4 * 13 \choose 3 * 3 * 13 \choose 2 = 4*286*3*78 = 267696

possibilities.

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Answer:

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Answer:

X=-7/3

Step-by-step explanation:

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