I believe it is yes. I think the right fraction is 84/1000. Let me know if this works and if not tell me
Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
Answer:
84
Step-by-step explanation:
Pretend her first 3 quizzes were all 83 since that was the average. So the 4th was 87.
83+83+83+87 = 336
Divide by the total number of quizzes because they are equally weighted.
336/4 = 84
Her new average score is 84.
Answer:
Step-by-step explanation:
factors of 84 = 2* 2 * 3 * 7
1) 21 *4 ===> 21 beads , 4 rows
2) 14 * 6 ===> 14 beads , 6 rows
3) 12 * 7 ===> 12 beads, 7 rows
