Need help answering this question for #30
2 answers:
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Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
Answer:
(A) 9.5 m/s
(B) 5.225 m
Explanation:
vertical height (h) = 4.7 m
horizontal distance (d) = 9.3 m
acceleration due to gravity (g) = 9.8 m/s^{2}
initial speed of the fish (u) = 0 m/s
(A) what is the pelicans initial speed ?
- lets first calculate the time it took the fish to fall
s = ut +
since u = 0
s =
t =
where a = acceleration due to gravity and s = vertical height
t = = 0.98 s
- pelicans initial speed = speed of the fish
speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s
initial speed of the pelican = 9.5 m/s
(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
vertical height = 1.5 m
pelican's speed = 9.5 m/s
- lets also calculate the time it will take the fish to fall
s = ut +
since u = 0
s =
t =
where a = acceleration due to gravity and s = vertical height
t = = 0.55 s
distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m
Answer:
a. μ 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ±(t*s)/sqrt(n)
where:
μ = is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.