Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
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 problem, we have that:

What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So



has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater
Answer:
2/5
Step-by-step explanation:
m=change in y/change in x
m=
m = 6-(4)/3-(-2)
m=2/3-(-2)
m= 2/3+2
m=2/5
37 * 48 = 3 * (c - 140 )+ 2 *(c - 140 ) + ( c - 140 ) =>
37 * 48 = 6 * ( c - 140 ) =>
37 * 48 / 6 = c - 140 =>
37 * 8 = c - 140 =>
296 = c - 140 =>
c = 296 + 140 =>
c = 436 ounces.
(4,-1) turns into (1,4) because the equation for that is (x,y)->(y•-1,x)