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daser333 [38]
3 years ago
7

The sum of two numbers is 100. The sum of 8 and the larger number is equal to 5 times the smaller number. What is the smaller nu

mber?
Mathematics
2 answers:
Finger [1]3 years ago
7 0

Answer:

The smaller number is 18.

Step-by-step explanation:

We have given that

The sum of two numbers is 100. The sum of 8 and the larger number is equal to 5 times the smaller number.

Let x and y are two numbers.

Let y is the larger number.

x+y  = 100      eq(1)

y+8 = 5x         eq(2)

From eq(1), we have

y  = 100-x

Putting above eq(3) into eq(2), we have

100-x+8 = 5x

108 = 5x+x

108 = 6x

x  = 108/6

x = 18

Hence, the smaller number is 18.

garik1379 [7]3 years ago
6 0

Answer:

The smaller number is 18

Step-by-step explanation:

Let the numbers be x and y where x is the smaller number; y\:>\:x.

The sum of the numbers is 100.

x+y=100...(1)

The sum of 8 and the larger number is equal to 5 times the smaller number.

y+8=5x

\Rightarrow y=5x-8...(2)

Put equation (2) into equation (1).

x+5x-8=100

x+5x=100+8

6x=108

x=18

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a) 0.4219 = 42.19% probability that one out of the next four cars needs oil.

b) 0.3115 = 31.15% probability that two out of the next eight cars needs oil.

c) 0.2581 = 25.81% probability that three out of the next 12 cars need oil.

Step-by-step explanation:

For each car, there are only two possible outcomes. Either they need oil, or they do not need it. The probability of a car needing oil is independent of any other car, which means that the binomial probability distribution is used 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.

One out of four cars needs to have oil added.

This means that p = \frac{1}{4} = 0.25

a. One out of the next four cars needs oil.

This is P(X = 1) when n = 4. So

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

P(X = 1) = C_{4,1}.(0.25)^{1}.(0.75)^{3} = 0.4219

0.4219 = 42.19% probability that one out of the next four cars needs oil.

b. Two out of the next eight cars needs oil.

This is P(X = 2) when n = 8. So

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

P(X = 2) = C_{8,2}.(0.25)^{2}.(0.75)^{6} = 0.3115

0.3115 = 31.15% probability that two out of the next eight cars needs oil.

c.Three out of the next 12 cars need oil.

This is P(X = 3) when n = 12. So

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

P(X = 3) = C_{12,3}.(0.25)^{3}.(0.75)^{9} = 0.2581

0.2581 = 25.81% probability that three out of the next 12 cars need oil.

6 0
3 years ago
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