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3 years ago

The sum of 5 consecutive integers is 505 what is the third number in the sequence

1 answer:

3 0

5 consecutive integers : x, x + 1, x + 2, x + 3, x + 4

x + x + 1 + x + 2 + x + 3 + x + 4 = 505
5x + 10 = 505
5x = 505 - 10
5x = 495
x = 495/5
x = 99

x + 1 = 99 + 1 = 100
x + 2 = 99 + 2 = 101 <=== 3rd number is 101
x + 3 = 99 + 3 = 102
x + 4 = 99 + 4 = 103

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b) 0.2977

c) 0.56

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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.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that $p = 0.13$

14 drivers are stopped

This means that $n = 14$

(a) None will have an invalid license.

This is $P(X = 0)$

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(b) Exactly one will have an invalid license.

This is $P(X = 1)$

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

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(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56$

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