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Scilla [17]
3 years ago
14

The sum of three numbers is 64. The second number is 3 more than the first. The third number is 11 less than twice the first. Fi

nd the numbers.
Mathematics
1 answer:
Cloud [144]3 years ago
8 0
The first number: x
The second number: x + 3
The third number 2 * x - 11
Sum of all numbers: 65

We can write it down as an equation:

65 = x + 3 + x + 2x - 11
65 = 4x + 3 - 11
65 = 4x - 8                         / + 8 (both sides)
73 = 4x                             / ÷ 6 (both sides_
x = 18,25

Doublecheck:

The first number: 18,25
The second number 18,25 + 3 = 21,25
The third number: 2 * 18,25 - 11 = 36,5 - 11 = 25,5
Sum of all numbers: 18,25 + 21,25 + 25,25 = 65, so it's correct :)
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According to national data, 5.1% of burglaries are cleared with arrests. A new detective is assigned to six different burglaries
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Answer:

26.95% probability that at least one of them is cleared with an arrest

Step-by-step explanation:

For each burglary, there are only two possible outcomes. Either it is cleared, or it is not. The probability of a burglary being cleared is independent of other burglaries. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

5.1% of burglaries are cleared with arrests.

This means that p = 0.051

A new detective is assigned to six different burglaries.

This means that n = 6

What is the probability that at least one of them is cleared with an arrest?

Either none are cleared, or at least one is. The sum of the probabilities of these events is 100% = 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1)

Then

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.051)^{0}.(0.949)^{6} = 0.7305

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7305 = 0.2695

26.95% probability that at least one of them is cleared with an arrest

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