A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The
foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder at time t seconds is given by the parametric equations (7 +2t, 0). The location of the top of the ladder will be given by parametric equations (0, y(t)). The formula for y(t)=√[625−(7+2t)²]. The domain of t values for y(t) ranges from 0 to 9.
Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
a) [0,2]
b) [2,4]
c) [6,8]
d) [8,9]
Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
a) [0,2]
b) [2,4]
c) [6,8]
d) [8,9]
1 answer:
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Answer:
1.05 ± 0.05 lbs
Step-by-step explanation:
Hi!
We can calculate this interval with the z-score of the 90% which is (by convention) 1.645
The interval is calculated as follows:
where x_m is the mean, σ the standar deviation and n is the number of samples:
replacing these values we get:
*rounded to the first decimal*
1.05 ± 0.05