The answer you are looking for is B
She’ll have 4 groups of 6 students and one group of five.
6x4= 24 students.
29-24= 5 students left over
Answer:
82.31% of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

Approximately what percentage of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter?
This is the pvalue of Z when X = 5.4 subtracted by the pvalue of Z when X = 4.2. So
X = 5.4



has a pvalue of 0.9842
X = 4.2



has a pvalue of 0.1611
0.9842 - 0.1611 = 0.8231
82.31% of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter
Answer:
18.0167≤x≤21.9833
Step-by-step explanation:
Given the following
sample size n = 10
standard deviation s = 3.2
Sample mean = 20
z-score at 95% = 1.96
Confidence Interval = x ± z×s/√n
Confidence Interval = 20 ± 1.96×3.2/√10
Confidence Interval = 20 ± (1.96×3.2/3.16)
Confidence Interval = 20 ± (1.96×1.0119)
Confidence Interval = 20 ± 1.9833
CI = {20-1.9833, 20+1.9833}
CI = {18.0167, 21.9833}
Hence the required confidence interval is 18.0167≤x≤21.9833
Answer:
32500 Days
Step-by-step explanation:
×
= 
We know that there are 24 hours in a day.
⇒24 Hours = 1 Day
therefore
hours =
Days