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
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.
I thinks it’s B because up it’s y and down its -y and right is X and left is -X
25% Because 50/200 is 0.25 which is 25%. Hope I helped.
Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
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:

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

Hence, the conditional probability is:

0.7873 = 78.73% probability that Mona was justifiably dropped.
A similar problem is given at brainly.com/question/14398287