Answer:
Y = Your age S = Sister's age
Y = 2S - 3
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
If the value of a is positive, the line slopes upwards.
First divide 3.7 by 0.006
Which is 616.666
The last 6 is the thousandths place. If it is 5 or bigger then the number to the left is rounded up. If its lower than 5 you round down
Since 6 is over 5 we tound up
616.67
Answer:
3101.8.
Keep Multiplying it by 1.12%, adding the result to the main number.