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Art [367]
3 years ago
6

Help me pleaseeeeeee

Mathematics
2 answers:
dusya [7]3 years ago
5 0
The answer should be C.
tiny-mole [99]3 years ago
5 0

Answer:

C

Step-by-step explanation:

18 x 5.85 = 105.3

105.3 is closest to the estimate of $120

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Please help 2 x ( 6 ÷ 2 + 8 ) - 4​
11111nata11111 [884]
2x(3+8)-4
2x(11)-4
22-4
18
6 0
2 years ago
PLZ HELP! A researcher wants to find out what brand of tomato sauce is most popular with people who work full-time. He samples s
arlik [135]

Answer:

No it is not a good sample because if a person works full time then he would be at work around that time.

Step-by-step explanation:

7 0
3 years ago
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.

3 0
3 years ago
What is the value of x? (There should be an image) ​
skad [1K]

Answer:

22

Step-by-step explanation:

6 0
2 years ago
What does "As Much" mean in math? I am doing homework and I don't want to fail so I really need help.
igor_vitrenko [27]
It means the same used in a question like this, is 25 as much as 25.
4 0
3 years ago
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