A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high IQ society. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. what score must a person have to qualify for Mensa? If required, round your answers to nearest whole number.£

130.81

Step-by-step explanation:

Given that :

Mean, μ = 100

Standard deviation, σ = 15

To obtain the upper 2% of scores :

We find the Zscore (value) of the upper 2% from the normal probability distribution table ;

Zscore corresponding to the area in the left of (1 - 0.02) = 2.054

Using this with the Zscore formula :

Zscore = (x - μ) / σ

2.054 = (x - 100) / 15

2.054 * 15 = x - 100

30.81 = x - 100

30.81 + 100 = x

x = 130.81

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .