Drown lemme know if this helps!
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
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
Answer:
https://www.pittsfordschools.org/cms/lib/NY02205365/Centricity/Domain/683/ANSWER%20KEY%20TO%20ALL%20HOMEWORK.pdf
Step-by-step explanation:
If this doesn’t work I’m sorry but that’s the link!
That is a 4 out of 10 chance that the second number is odd, as well as the first.