Answer:
The critical value for a 98% CI is z=2.33.
The 98% confidence interval for the mean is (187.76, 194.84).
Step-by-step explanation:
We have to develop a 98% confidence interval for the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day.
We know the standard deveiation of the population (σ=21.5 min.).
The sample mean is 191.3 minutes, with a sample size n=200.
The z-value for a 98% CI is z=2.33, from the table of the standard normal distribution.
The margin of error is:

With this margin of error, we can calculate the lower and upper bounds of the CI:

The 98% confidence interval for the mean is (187.76, 194.84).
The perimeter is 18.85 feet
16.4 because the number after 4 is 0 so it stays the same.
We observe that 14 is 4 more than twice 5.
The scooter's rate is 14 mph.
The bicycle's rate is 5 mph.
_____
time = distance/rate
14/(2b +4) = 5/b . . . . . . let b represent the bicycle's speed in mi/h
14b = 10b +20 . . . . . . . .cross multiply
4b = 20 . . . . . . . . . . . . . subtract 10b
b = 5 . . . . . . . . . . . . . . . divide by 4. Same answer as above.
Answer:
60
Step-by-step explanation:
write 120 like this: 120/2
Then just multiply the numerators.
Which gets you 120/2.
Then change it it to a mixed fraction which is 60.