Answer:
The value is c = 21.1445.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb.
This means that 
What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?
This 1 added to the value of X for the 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.




1 + 20.1445 = 21.1445
The value is c = 21.1445.
The answer to your question is 17/90:D
Answer:
Thanks for the free point
Answer:
22
Step-by-step explanation:
5*5=25
25-3=22
Answer:
AOX = 30
YOD = 12
DOB = 48
Step-by-step explanation:
All of these angles are supplementary which means they all add up to 180:
x + 18 + 90 + x + 4x = 180
6x + 108 = 180
6x = 180 - 108
6x = 72
x = 12
m<AOX:
x + 18
12 + 18
30
m<YOD:
x
12
m<DOB:
4x
4(12)
48