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:
-25/9
Step-by-step explanation:
-5/6 : 3/10
-->(cross multibly)
=-5 *10 : 6*3
=-50/ 18
=-25/9
Answer:
y = 2
Step-by-step explanation:
not sure but should be it
HOW TO SOLVE:
1) use x as a variable
2) the starting equation is:
4x + 3x + 5x = 18
3) add all the like terms on the left side of the equation:
12x = 18
4) Divide by 12 on both sides:
x=1.5
5) Find the lengths of the sides:
- 4x = 4(1.5) = 6
- 3x = 3(1.5) = 4.5
- 5x = 5(1.5) = 7.5
SO, the lengths of the sides of this triangle are: 6, 4.5, and 7.5