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:
3
1
if your estimating to whole numbers the 3 1/8 is 3 and 1 2/5 is 1
Step-by-step explanation:
<span>The number can be 102, 108, or 114. They can all be formed that way.
32 + 34 + 36 = 102
34 + 36 + 38 = 108
36 + 38 + 40 = 114
</span>The n<span>ext one is 120 . . . . . 38 + 40 + 42</span>
Answer:
see below
Step-by-step explanation:
— 7x — бу = — 15
To find the x intercept set y = 0 and solve for x
-7x - 0 = -15
-7x= -15
Divide by -7
-7x/-7 = -15/-7
x = 15/7
The x intercept is (15/7,0)
To find the y intercept set x = 0 and solve for y
o— бу = — 15
— бу= -15
Divide by -6
— бу/-6= -15/-6
у= 5/2
The y intercept is (0,5/2)
Answer:
1/72
Step-by-step explanation:
Not entirely sure but i do hope it helps
