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pychu [463]
2 years ago
14

I need the steps too

Mathematics
1 answer:
Musya8 [376]2 years ago
6 0

Answer:

x =10

Step-by-step explanation:

7(3x-1)-(x+5)

21x-7-x+5

20x-2=-52

x=20÷2

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Sholpan [36]

You need two equations to solve a two-variable equation.

7 0
3 years ago
After the holidays, Christmas decorations are marked down 60%! If Mr. Buchser wanted to buy an inflatable snowman for the front
Dmitry_Shevchenko [17]
$160* 0.6= 96
160-96= $64
The snowman cost $64
3 0
2 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
I NEED HELP!!! PLZ HELP!!!!!!!!!
Digiron [165]

Answer:

30.06

Step-by-step explanation:

The question states WHAT DID THEY PAY BEFORE INSTALATION CHARGES.

So 30.06$

588.34-555.28=30.06

3 0
3 years ago
Which of the numbers 285, 629, 5490, 15708, 43695 are divisible by 3 but not by 9?
Ne4ueva [31]

Answer:

285

Step-by-step explanation:

285÷3=95

285÷9=31.6666

3 0
2 years ago
Read 2 more answers
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