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bekas [8.4K]
3 years ago
7

Edith is x years old. Her sister,Madison is six years older than her. Their mother is twice as old as Madison. Their aunt, Eliza

beth is x years older than their mother. Write and simplify an expression that represents Elizabeth's age in years.
Mathematics
1 answer:
DENIUS [597]3 years ago
5 0

Answer:

Step-by-step explanation:

Edith is x years old.

Her sister,Madison is six years older than her. This means that the age of Madison is x + 6

Their mother is twice as old as Madison. This means that the age of their mother is expressed as

2(x + 6)

= 2x + 12

Their aunt, Elizabeth is x years older than their mother. This means that the expression that represents Elizabeth's age in years is

2x + 12 + x

= 3x + 12

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Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo
natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

So

P(B) is the probability of the person having cancer, so P(B) = 0.009

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So P(A/B) = 0.91

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

There is a 12.13% probability that the person actually does have cancer.

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