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mixas84 [53]
3 years ago
8

Please help me quick :(

Mathematics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

I don't know how helpful I'll be, sorry :(

Step-by-step explanation:

So the scale factor is 1/4 reduction because the triangle P'Q'R' is smaller than the original. When reflecting the triangle P'Q'R' over the Y axis, you'll have P"Q"R". The graphing might be a little confusing but it's just a vertical reflection.

P" = (-1,0)

Q"= (0,-1)

R"= (2,-1)

Triangles PQR and P"Q"R" are not congruent because P"Q"R" has had a 1/4 reduction and a vertical reflection.

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Which terms could have a greatest common factor of 5m2n2?.
ozzi

Answer:

a common factor would be 5m^4n^3 or 15m^2n^2

Step-by-step explanation:

not sure what terms they want you to choose from but anything that starts with a number (5,10,15,) and "m" would any number 5 would go into

3 0
3 years ago
Among 420 randomly selected employees at a company, the mean number of hours of overtime worked per month is 10 hours and the st
olganol [36]
To solve for the margin of error:

margin of error = critical value x standard error

To compute for the critical value:
1. compute alpha: α= 1 - (confidence level/100) = 1 - .99 = 0.01
2. Find the critical probability: p = 1 - α/2 = 1 - (0.01/2) = 0.995
3. Look for the z score at the z table, in this case it is 2.58. This is your critical value.

To compute for the standard error, the formula is:
standard error = standard deviation (σ) divided by √n = 1.6 / √420 = 0.0781

Now plug in the following in the formula:

margin of error: 2.58 * 0.0781  = 0.2014 or 0.20

The answer is letter C.
3 0
3 years ago
Read 2 more answers
Which equation represents the graph below?
Art [367]
I thinks it’s B because up it’s y and down its -y and right is X and left is -X
6 0
3 years ago
There are 200 employees in your company Fifty of them do not have company health insurance what percentage of the employees do n
givi [52]
25% Because 50/200 is 0.25 which is 25%. Hope I helped.
6 0
4 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
3 years ago
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