Answer:
2.28% probability that a person selected at random will have an IQ of 110 or higher
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 higher?
This is 1 subtracted by the pvalue of Z when X = 110. So



has a pvalue of 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or higher
Answer:
#3. 16a + 24b
Step-by-step explanation:
P= 2( length plus width).
98 degrees.
30+52=82
180-82=98
Simplifying
h(t) = -1t2 + -2t + 30
Multiply h * t
ht = -1t2 + -2t + 30
Reorder the terms:
ht = 30 + -2t + -1t2
Solving
ht = 30 + -2t + -1t2
Solving for variable 'h'.
Move all terms containing h to the left, all other terms to the right.
Divide each side by 't'.
h = 30t-1 + -2 + -1t
Simplifying
h = 30t-1 + -2 + -1t
Reorder the terms:
h = -2 + 30t-1 + -1t
C = 5/9(F - 32)...multiply both sides by 9/5, eliminating the 5/9 on the right
9/5C = F - 32...add 32 to both sides
9/5C + 32 = F <===