Answer:
The probability that the next failure will not occur before 30 months have elapsed is 0.0454
Step-by-step explanation:
Using Poisson distribution where
t= number of units of time
x= number of occurrences in t units of time
λ= average number of occurrences per unit of time
P(x;λt) = e raise to power (-λt) multiplied by λtˣ divided by x!
here λt = 25
x= 30
P(x= 30) = 25³⁰e⁻²⁵/ 30!
P (x= 30) = 8.67 E41 * 1.3887 E-11/30! (where E= exponent)
P (x=30) = 1.204 E31/30!
Solving it with a statistical calculator would give
P (x=30) = 0.0454
The probability that the next failure will not occur before 30 months have elapsed is 0.0454
Let x represent the number of type A table and y represent the number of type B tables.
Minimize: C = 265x + 100y
Subject to: x + y ≤ 40
25x + 13y ≥ 760
x ≥ 1, y ≥ 1
From, the graph the corner points are (20, 20), (39, 1), (30, 1)
For (20, 20): C = 265(20) + 100(20) = $7,300
For (39, 1): C = 265(39) + 100 = $10,435
For (30, 1): C = 265(30) + 100 = $8,050
Therefore, for minimum cost, 20 of type A and 20 of type B should be ordered.
I’m pretty sure the answer is test
The rejection region is give by
where the test statistics is given by
i.e.
Thus,
Using the statistical table, the level of the test is 0.04.
I believe this would take the form of an exponential
equation:
A = Ao (1 + r)^t
where A is final population, Ao is initial population, r
is rate of growth and t is time
A / Ao = (1 + r)^t
log A / Ao = t log (1 + r)
t = (log A / Ao) / log (1 + r)
t = [log (1000 / 550)] / log (1.075)
t = 8.27 years
SO the answer is B) about 9 years
