Answer:
(10.78483, 12.61517)
Step-by-step explanation:
It is given that the genera mangers of few hotels were sent some questionnaires for conducting a study for the career paths in the major hotel chains of the United States.
Number of hotels = 160
Number of response received = 103
The average number of years these general mangers was in their current hotels,
= 11.7 years
Confidence Interval, CI = 0.99
Therefore,
a = 0.01, |Z(0.005)| (from standard normal table)
∴ 99% of CI = 


Let z = sin(x).
.. z^2 -2z = 2
.. z^2 -2z +1 = 3
.. (z -1)² = 3
.. z -1 = ±√3
.. z = 1 -√3 . . . . . the positive solution is extraneous
sin(x) = 1 -√3
Answer:
27 and 23
Step-by-step explanation:
x + y = 50
x - y = 4
2x = 54
x = 27
y = 23
T = 40000 + 40000(1.03) + 40000(1.03)^2 + ... + 40000(1.03)^15
Multiply the above equation by 1.03 giving
1.03(T) = 40000(1.03) + 40000(1.03)^2 + 40000(1.03)^3 + ... + 40000(1.03)^15 + 40000(1.03)^16
Now subtract the first equation from the second to give
0.03T = 40000(1.03)^16 - 40000
0.03T = 40000(1.03^16 - 1)
0.03T = 24188.26
T = 24188.26 / 0.03 = $806275.25