Haiii.. i'm from indonesia..
thanks for point:)
Answer:
The 99% confidence interval would be given by (0.054;0.154)
. So we are confident at 99% that the true proportion of people that they did work at home at least once per week is between 0.054 and 0.154
Step-by-step explanation:
For this case we can estimate the population proportion of people that they did work at home at least once per week with this formula:

We need to find the critical value using the normal standard distribution the z distribution. Since our condifence interval is at 99%, our significance level would be given by
and
. And the critical value would be given by:
The confidence interval for the mean is given by the following formula:
If we replace the values obtained we got:
The 99% confidence interval would be given by (0.054;0.154)
. So we are confident at 99% that the true proportion of people that they did work at home at least once per week is between 0.054 and 0.154
Answer: x is 20
Step-by-step explanation:
This is ratio
a : c = b : x and d : unknown
a : c = 9 : 12 = 3 : 4 which is 9/12
b : x = 15 : x which is 15 / x
So the proportion will be :
3/4 = 15/x
So Cross multiplying
3x = 90
Divide both side by 3
x = 90/3
x = 20
<span>-18, -13, -8, -3,…
-18 + 5 = -13
-13 + 5 = -8
-8 + 5 = -3
-3 + 5 = 2
2 + 5 = 7
7 + 5 = 12
Therefore, the next three terms are 2, 7, 12
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