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:
S = v1 t1 = 7 t1 traveling downstream
S = v2 t2 = 5 t2 traveling upstream
7 t1 = 5 t2
7 (6 - t2) = 5 t2 since t1 + t2 = 6
42 - 7 t2 = 5 t2
t2 = 42 / 12 = 3.5 hrs so t1 = 2.5 hrs
S = 7 t1 = 7 * 2.5 = 17.5 mi
Also, S = 5 t2 = 5 * 3.5 = 17.5 mi
Answer:
x = 6
Step-by-step explanation:
7 x = 56-14
7 x = 42
7 x/7 to remain with x only = 42
then x = 6
Answer:
2 terms
Step-by-step explanation:
Terms are nothing but the mixtures of co-effecients and their variables..
Here, 9a is one term while 11p is the other
Answer:
(4, 1)
Step-by-step explanation:
8 units to the right = x + 8, so -4 + 8 = 4
4 units up = y + 4, so -3 + 4 = 1