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daser333 [38]
3 years ago
9

a brother is 12 years older than his sister. four years ago, the brother was 3 times as old as the sister is now. how old is the

broter now? a 12 b 14 c 16 d 18
Mathematics
2 answers:
Yakvenalex [24]3 years ago
8 0

Answer:

C. 16

Step-by-step explanation:

Let the brother's age be x.

Let the sister's age be y.

x = y + 12

x - 4 = 3y

Put x as y + 12 in the second equation.

y + 12 - 4 = 3y

y + 8 = 3y

8 = 3y - y

8 = 2y

y = 4

Put y as 4 in the first equation.

x = 4 + 12

x = 16

The brother is 16 years old and the sister is 4 years old.

jasenka [17]3 years ago
6 0

Answer:

C. 16

Step-by-step explanation:

Brother - b years now.

Sister - s years now.

A brother is 12 years older than his sister.

b - s = 12

(b - 4) - brother 4 years ago.

Four years ago, the brother was 3 times as old as the sister is now.

b - 4 = 3s

b - s = 12 ----> s = b - 12

b - 4 = 3s

b - 4 = 3(b - 12)

b - 4 = 3b - 36

36 - 4 = 3b - b

32 = 2b

b = 16

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Answer:

1001.

Step-by-step explanation:

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A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb
Maksim231197 [3]

Answer:

a) P(X = 1) = 0.38742

b) P(X = 3) = 0.05740

c) P(X = 9) = 0.00000

d) P(X \geq 5) = 0.00163

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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

There are 10 containers, so n = 10.

A food-packaging apparatus underfills 10% of the containers, so p = 0.1.

a) This is P(X = 1)

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

b) This is P(X = 3)

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

c) This is P(X = 9)

P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000

d) This is P(X \geq 5).

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

P(X < 5) + P(X \geq 5) = 1

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

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

P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742

So

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837

Finally

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163

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