31.5 times 1.8.
=31.5×.8+31.5×1
=25.2+31.5
=56.7
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:
what i have lerned in this module
Answer:
one-quarter times the number of pears, plus four and one-half is forty-four and one-half
Step-by-step explanation:
The total weight (44.5 lb) will be the sum of the empty weight (4.5 lb) and the number of pears multiplied by the pear's weight. That can be expressed as ...
one-quarter times the number of pears, plus four and one-half is forty-four and one-half
