You are climbing in the high sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of this cliff, you drop a rock from the top and 7: 40 s later hear the sound of it hitting the ground at the foot of the cliff. ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s?

Step-by-step explanation:

The distance that the rock has traveled to reach the ground (also, the height of the cliff) can be calculated using:

is our incognit

is the starting position of the rock, we'll define it as 0

is the starting velocity of the rock, it is 0 since it starts from rest

g is the acceleration of gravity

is the time it took the rock reaching the ground

Also, we can model the "movement" of sound as it follows:

where:

is the velocity of sound :330m/s

is the time it took the sound travelling to the top of the cliff

using the first and the second equation, and the fact that we get to:

(we get to values for tr since the equation is quadratic, the correct one is the positive)

Now that we have we can use it to calculate and determine the height of the cliff:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;