Answer;
The relevant probability is 0.136 so the value of 56 girls in 100 births is not a significantly high number of girls because the relevant probability is greater than 0.05
Step-by-step explanation:
The complete question is as follows;
For 100 births, P(exactly 56 girls = 0.0390 and P 56 or more girls = 0.136. Is 56 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question? Consider a number of girls to be significantly high if the appropriate probability is 0.05 or less V so 56 girls in 100 birthsa significantly high number of girls because the relevant probability is The relevant probability is 0.05
Solution is as follows;
Here. we want to know which of the probabilities is relevant to answering the question and also if 56 out of a total of 100 is sufficient enough to provide answer to the question.
Now, to answer this question, it would be best to reach a conclusion or let’s say draw a conclusion from the given information.
The relevant probability is 0.136 so the value of 56 girls in 100 births is not a significantly high number of girls because the relevant probability is greater than 0.05
List the factors:
21: 1, 3, 7, 21
40: 1, 2, 4, 5, 8, 10, 20, 40
They only have 1 in common.
So 1 is the greatest common factor.
Please, write "x^3" for "the cube of x," not "x3." "^" denotes exponentiation.
Then you have g(x) = x^3 - 5 and (I assume) h(x) = 2x - 2.
1) evaluate g(x) at x = -2: g(-2) = (-2)^3 - 5 = -8 - 5 = -13
2) let the input to h(x) be -13: h(-13) = 2(-13) - 2 = -28 (answer)
Answer:
-1x - 3 or -x - 3
Step-by-step explanation:
1/5 (5x - 15) - 2x =
1x - 3 - 2x =
1x - 2x - 3 =
-1x - 3
Answer:
It depends on what q is.
Step-by-step explanation:
Is d is less than q then q will be greater.
