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kolbaska11 [484]
3 years ago
9

22 points + brainliest! A fair die with sides labeled 1 through 6 is rolled two times. The values of the two rolls are added tog

ether. The sum is recorded as the outcome of a single trial of a random experiment. Compute the probability that the sum is 9.

Mathematics
2 answers:
murzikaleks [220]3 years ago
8 0

Answer:

P(9) = 1/9

Step-by-step explanation:

From the contingency table, we see that 9 appears 4 times out of the 36 possible outcomes, therefore the probability of having a sum of 9 is

P(9) = 4/36 = 1/9

ki77a [65]3 years ago
3 0

If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.

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Answer:

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Step-by-step explanation:

Starting from the theory we have the following equation:

fi*P(x

Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:

P( z \leq \frac{c-1-12}{3.5}) =0.99/fi

solving for "c", knowing that fi is a tabulating value:

\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155

therefore the value of c is equal to 21.16

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3 years ago
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Answer:

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8 0
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