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
It’s less than cuz 2/10 is less than 3/10 (<)
<em>Hey</em><em>!</em><em>!</em><em>!</em>
<em>here</em><em>'s</em><em> </em><em>your</em><em> </em><em>answer</em>
<em>X+</em><em>1</em><em>2</em><em>8</em><em>=</em><em>1</em><em>8</em><em>0</em><em>(</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>angle</em><em> </em><em>in</em><em> </em><em>straight</em><em> </em><em>line</em><em>)</em>
<em>or</em><em>,</em><em>X=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>1</em><em>2</em><em>8</em>
<em>X=</em><em>5</em><em>2</em><em> </em><em>degree</em><em>.</em>
<em>So</em><em> </em><em>the</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>X </em><em>is</em><em> </em><em>5</em><em>2</em><em> </em><em>degree</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Combine like terms and then simplify
Answer:
Step-by-step explanation:
DE=DG
8s-97=3s-22
5s-97=-22
5s=75
s=15
