Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
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
(17-7)*2=weekly allowance
so her weekly allowance is 20
105,264 Lego’s in a case (please give brainlist)
