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olga nikolaevna [1]
3 years ago
15

Suppose the probability that a randomly selected​ man, aged 55​ - 59, will die of cancer during the course of the year is StartF

raction 300 Over 100 comma 000 EndFraction . How would you find the probability that a man in this age category does NOT die of cancer during the course of the​ year?
Mathematics
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

The probability that a man in this age category does NOT die of cancer during the course of the​ year is 0.997.

Step-by-step explanation:

Suppose the probability of an event occurring is P_{i}.

The probability of the given event not taking place is known as the complement of that event.

The probability of the complement of the given event will be,

1 - P_{i}

In this case an events <em>X</em> is defined as a man, aged 55​ - 59, will die of cancer during the course of the year.

The probability of the random variable <em>X</em> is:

P (X) = \frac{300}{100000}=0.003

Then the event of a man in this age category not dying of cancer during the course of the​ year will be complement of event <em>X, </em>denoted by<em> X</em>'.

The probability of the complement of event <em>X</em> will be:

P(X')=1-P(X)

          =1-0.003\\=0.997

Thus, the probability that a man in this age category does NOT die of cancer during the course of the​ year is 0.997.

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If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos
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Answer:

The least possible result is <em>-10</em>.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

A(B-C) = 4 \times (5-6) = 4 \times -1 = -4

Let A = 4, B = 6 and C = 5

A(B-C) = 4 \times (6-5) = 4 \times 1 = 4

Let A = 5, B =4 and C = 6

A(B-C) = 5 \times (4-6) = 5 \times -2 = -10

Let A = 5, B = 6 and C = 4

A(B-C) = 4 \times (6-4) = 4 \times 2 = 8

Let A = 6, B = 4 and C = 5

A(B-C) = 6 \times (4-5) = 6 \times -1 = -6

Let A = 6, B = 5 and C = 4

A(B-C) = 6 \times (6-5) = 6 \times 1 = 6

Summarizing the above values in the form of a table:

\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\  4 & 6 & 5 & 4\\  5 & 4 & 6 & -10\\  5 & 6 & 4 & 10\\  6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}

So, the least possible result is <em>-10</em>.

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3 years ago
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