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bonufazy [111]
3 years ago
5

8. (0,9) is an example of what?

Mathematics
1 answer:
Law Incorporation [45]3 years ago
4 0
B, y-intercept, because since the x coordinate for the ordered pair is 0, it is showing y-intercept
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PLS HELP
Charra [1.4K]

Answer:  -418, 728, 446, -429, 1159, -696

Step-by-step explanation:

1. n = -199 - 219 = \boxed{-418}.

2. m = 438 - (-290) = 438 + 290 = \boxed{728}.

3. 128 - (-318) = 128 + 318 = \boxed{446}.

4. The sum of 429 and 429 is 429+429 = 858. We need to subtract this from 429, so 429 - 858 = \boxed{-429}.

5. The sum of 510 and 319 is 510+319 = 829. Subtracting -330, we get 829 - (-330) = 829 + 330 = \boxed{1159}.

6. Call the unknown number "\star". Then \star + 298 = -398, so \star = -398 - 298 = \boxed{-696}. (Think about it.)

4 0
3 years ago
Lydia invested $8,300 in an account paying an interest rate of 2.2% compounded
Dennis_Churaev [7]

Answer:

1245

Step-by-step explanation:

5 0
3 years ago
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What is this please? i'll give u brainliest
Zolol [24]

Answer:

i dont know i just want points

Step-by-step explanation:

>;)

4 0
3 years ago
What is the product?
ad-work [718]

Answer is C, just multiply and expand.

7 0
3 years ago
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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.

3 0
3 years ago
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