He takes a random sample of 49 recent charterholders and computes a mean salary of $172,000 with a standard deviation of $35,000
1 answer:
Answer:
the 90% of confidence intervals for the average salary of a CFA charter holder
(1,63,775 , 1,80,000)
Step-by-step explanation:
<u>Explanation</u>:-
random sample of n = 49 recent charter holders
mean of sample (x⁻) = $172,000
standard deviation of sample( S) = $35,000
Level of significance α= 1.645
<u> 90% confidence interval</u>
on calculation , we get
(1,63,775 , 1,80,000)
The mean value lies between the 90% of confidence intervals
(1,63,775 , 1,80,000)
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Answer:
a) The tree is 32.98 feet tall.
b) The bird is 50.21 feet high in the air.
c) The Bird is 17.23 feet far above the Tree.
Step-by-step explanation:
From the attached diagram, please familiarize yourself with the statement drawn out and the points drawn out for better understanding.
Note that Line CD is the ground level and Line QE is the Wendy's eyes height.
1) <em>How tall is the tree</em> (Line BC)
line BC (tree height) => QC + QB
Line ED = QC, ∴ QC = <em>5 ft</em>
Since <BEQ = 25° and CD = QE, ∴ QE = 60ft all in triangle BEQ
Therefore Tan 25° = BQ/60
BQ = 60 (0.4663)
BQ = <em>27.98 ft</em>
tree height => QC + QB = 5 ft + 27.98 ft ==> 32.98 ft
2) <em>How high is the bird in the air?</em> (Line AC)
line AC (bird height) => QC + QA
Line ED = QC, ∴ QC = <em>5 ft</em>
Since <AEQ = 37° and CD = QE, ∴ QE = 60ft in triangle AEQ
Therefore Tan 37° = QA/60
QA = 60 (0.7536)
QA = <em>45.21 ft</em>
bird height => QC + QA = 5 ft + 45.21 ft ==> 50.21 ft
3) <em>How far above the tree is the bird?</em>
The difference between the bird height and the tree height = Line AC - Line BC
= 50.21 ft - 32.98 ft
= 17.23 ft
