Answer:
Unbiased
Step-by-step explanation:
If b^ is equal to B this means that it is an unbiased estimator. When there is an absence of bias, we have an unbiased estimator. As an unbiased estimator it gives accurate information most of the time. The result it gives is not over estimated and also it is not underestimated.
Expected value = true value
Parameter estimates are correct on average
Thank you
Answer:
The zeros are 8 and -10, all u have to do is substitute x for those values, factor it, or graph it.
Answer:
6
Step-by-step explanation:
(x + 9) * (x - 2) = 60
x^2 - 2x + 9x - 18 = 60
x^2 + 7x - 18 = 60
x^2 + 7x = 78
FACTORING:
x(x + 7) = 78
x = 6
Hello from MrBillDoesMath!
Answer:
See Discussion section below
Discussion:
2x^2 + 5x + 3 = (2x+3)(x+1) which is Choice 3
3x^2+10x+4 =-1/3 (-3 x + sqrt(13) - 5) (3 x + sqrt(13) + 5)
I don't think any of the provided choices are right!
8x^2+10x-3 = (2x+3)(4x-1) which is Choice 2
6x^2-7x-4 = -1/24 (-12 x + sqrt(145) + 7) (12 x + sqrt(145) - 7)
I don't think any of the provided choices are right!
2x^2+x-28 = (2x-7)(x+4) which is Choice 2
Thank you,
MrB
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

Find the probability that the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then



has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.