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Salsk061 [2.6K]
3 years ago
15

The ages of three siblings are consecutive integers. The sum of their ages is 42. Calculate their ages. *

Mathematics
1 answer:
stealth61 [152]3 years ago
8 0

Answer:

13, 14, 15

Step-by-step explanation:

Let x = age of one of the siblings

Since the age of the siblings are consecutive integers, the age of the next sibling would be x + 1 and the age of the third sibling would be x + 2

x +( x +1) + (x +2) = 42

3x + 3 = 42

collect like terms and solve for x

x = 13

The age of the first sibling is 13

The age of the second sibling is x + 1 = 13 + 1 = 14

The age of the third sibling is x + 2 = 13 + 2 = 15

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Answer:

x = 0

Step-by-step explanation:

x^3-7x^2-x =0\\

Take out common factor

x(x^2-7x-1) = 0

The factor inside the parenthesis cannot be factored

Since there is a common factor that we pulled out of the original equation,

set that x equal to zero and solve

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A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra
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Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that N = 6

4 blue, which means that k = 4

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that N = 6

2 yellow, which means that k = 2

Sample of 2, which means that n = 2

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that P(A) = 0.4 + 0.0667 = 0.4667

Probability of both being the same color and blue:

40% probability that both are blue, which means that P(A \cap B) = 0.4

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

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Answer:

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Step-by-step explanation:

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