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Elza [17]
2 years ago
10

C=-3. a=2 b=4 (2)2(3c+7)

Mathematics
1 answer:
Paha777 [63]2 years ago
4 0

I think the answer is -18 + 14

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PLEASE HELPP!! ILL MARK YOU BRAINLIEST!! :)
devlian [24]

Answer:

1) is 40%

2) is 4,307

3) is 35

I think it's correct you can double-check. I apologize if it's wrong.

Step-by-step explanation:

8 0
2 years ago
In how many ways can a class of 10 students be assigned 1 a, 2 b;s
nalin [4]
2.b..........................
6 0
3 years ago
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Joshua has $24.He used 5/8of his money.How much money did he spend
zhenek [66]
So 5/8 of $24
5/8 * 24
15

Joshua spent $15.

Hope this helps :)
5 0
3 years ago
Any athlete who fails the Enormous State University's women's soccer fitness test is automatically dropped from the team. Last y
castortr0y [4]

Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Fail the test.
  • Event B: Unfit.

The probability of <u>failing the test</u> is composed by:

  • 46% of 37%(are fit).
  • 100% of 63%(not fit).

Hence:

P(A) = 0.46(0.37) + 0.63 = 0.8002

The probability of both failing the test and being unfit is:

P(A \cap B) = 0.63

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287

3 0
2 years ago
Brainliest to who ever is correct
neonofarm [45]

Answer:  D i hope

Step-by-step explanation: your welcome

8 0
2 years ago
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