Answer:
25a^2-49
Step-by-step explanation:
1)use FOIL
F=first
O=out
I=inside
L=last
5a*5a+5a*-7+7*5a*7*-7
25a^2-35a+35a-49
25a^2-49
Hope this helps!
Answer:
- 1. 5 as decimal
-3 / 2 as a fraction
Step-by-step explanation:
Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.
Number of sample, n = 41
Mean, u = 43.1 hours
Standard deviation, s = 5.91 hours
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
43.1 ± 1.645 × 5.91/√41
= 43.1 ± 1.645 × 0.923
= 43.1 ± 1.52
The lower end of the confidence interval is 43.1 - 1.52 =41.58
The upper end of the confidence interval is 43.1 + 1.52 =44.62
Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
