R S + 14(S) R=6 S=1/4
(6)(1/4) + 14(1/4)
1.5 + 14(1/4)
1.5 + 3.5
5
Answer:
3x
Step-by-step explanation:
Answer:(a) margin error = 2.4%
(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.
(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%
Step-by-step explanation:
Using the proportion formulae
Margin error = z √p(1-p)/n
n= 1170, p = 53% = 0.53, 1-p = 0.47
and the z value at 90% C.I = 1.645
M error= 1.645 √0.53×0.47/1170
Margin error = 0.024 = 0.024 ×100
Margin error = 2.4%
53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%
In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.
7/20=2.857142857142857 divide 20 by 7
We subsituete
9:6=6:4
treat as fraction
9/6=6/4
9/6=3/3 times 3/2
6/4=2/2 times 3/2
3/2=3/2
this is TRUE
