Answer:
a) The equation is:
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b) The 98% confidence interval = (5.62784, 6.37216)
Step-by-step explanation:
a. Write down the equation you should use to construct the confidence interval for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b. Determine a 98% confidence interval estimate for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
Mean = 6 days
Standard deviation = 1.6 days
Number of samples = 100
Z score of 98% confidence interval = 2.326
Confidence interval = 6 ± 2.326 × 1.6/√100
= 6 ± 2.326 × 1.6/10
= 6 ± 0.37216
= 6 - 0.37216
= 5.62784
6 + 0.37216
= 6.37216
Therefore, the 98% confidence interval = (5.62784, 6.37216)
Answer:
.53
Step-by-step explanation:
Well all you gotta do is basically just divide so it will be 2.65 divide by 5 which would equal the answer which is .53 for all sides
Answer:
d = 10/72
Step-by-step explanation:
c and d vary inversely
c = k/d
Where,
k = constant of proportionality
d = 2/9 when c = 5
c = k/d
5 = k ÷ 2/9
5 = k × 9/2
5 = 9k/2
Cross product
5*2 = 9k
10 = 9k
k = 10/9
c = k/d
c = 10/9 ÷ d
c = 10/9 × 1/d
c = 10/9d
find d when c = 8
c = 10/9d
8 = 10/9d
Cross product
8*9d = 10
72d = 10
d = 10/72
Ttps://www.palmbeachstate.edu/prepmathlw/Documents/translatingkeywords.pdf
Try seeing if this link works it should help
