I came up with 114.48. i think you meant bill is $90, not o0%
Error = 1/2(Maximum range - Minimum range) = 1/2(3.2-1.1) = % = 0.0105
If p is the point estimator;
Then,
Minimum range = p - Error
Maximum range = p + Error
Solving one of the two expressions,
0.011 = p - 0.0105 => p = 0.011+0.0105 = 0.0215 = 2.15%
Answer:
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that the selection of the random pages will contain at least two errors is 0.2644
Step-by-step explanation:
From the information given:
Let q represent the no of typographical errors.
Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let be the random variable that follows a Poisson distribution, then mean
and the mean that the random selection of 50 pages will contain no error is
∴
Pr(q =0) = 0.368
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that 50 randomly page contains at least 2 errors is computed as follows:
P(X ≥ 2) = 1 - P( X < 2)
P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2
P(X ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644
Answer:
66 yards squared
Step-by-step explanation:
12x11/2=66