H=-4.9t^2+25t
0=t(-4.9t+25)
T=25÷4.9
T= approx 5.1 which is closes to 5.0 So answer will be 5.0
A=-112 bcs -216 ft is down from sea level and -104 is also down from sea level bcs both values are in minus but -104 is higher than other so you can minus it from -216 the difference is the answer
b=-109
Answer:

Step-by-step explanation:
The formula relating distance (d), speed (s), and time (t) is
d = st
1. Calculate the distance
d = 269 s × 4.6 m·s⁻¹ = 1240 m
2.Calculate the track radius
The distance travelled is the circumference of a circle

Answer:

Step-by-step explanation:
For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.
We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The confidence interval for a proportion is given by this formula
For the 98% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And the margin of error would be:
