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Alexxx [7]
3 years ago
12

Plz help ITS NOT A LOT !!!!!! Plz help

Mathematics
1 answer:
padilas [110]3 years ago
3 0

Answer: Actually, you must use the Pythagorean Theorem!

a squared plus b squared equals c squared. So if you substitute in the values, a is 2 units and b is three units. 2 squared is 4 and 3 squared is 9.

4 + 9 = 13 so the answer is the square root of 13. (√13)

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The system of equations cx + 3y = c – 3 and 12x + cy = c will have infinitely many solutions if the value of c is
irinina [24]

Answer:

6

Step-by-step explanation:

1. re-write the given system:

cx+3y=c-3; => y= -cx/3 +(c-3)/3

12x+cy=c; => y= -12x/c+c/12

2. according the condition the rule for the parallel graphs is:

-c/3= -12/c

3. to calculate the unknown 'c':

c²=36; ⇔c=±6

7 0
3 years ago
Read 2 more answers
I need help please help. 2 Questions.
Colt1911 [192]

Answer:

Step-by-step explanation:

8 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
1/3 of 90 = 2/3 of ?
rewona [7]
Answer=45
1/3 of 90= 30
5 0
3 years ago
Which law would you use to simplify the expression?
vivado [14]

Answer:

A. quotient of Powers

Step-by-step explanation:

hope it helps

8 0
3 years ago
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