The average test scores for a particular test in algebra 2 was 84 with a standard deviation of 5. what percentage of the students scored higher than 89%? explain your answer.

Percentage of the students scored higher than 89% =  65.87%

Step-by-step explanation:

We need to find the percentage of students who scored higher than 89% given mean = 84 and standard deviation = 5.

We can find the percentage of students who scored higher than 89% by using z-score.

The formula for z-score is:

μ= mean = 84

σ=standard deviation = 5

x= random number = 89

Putting these values, we get

Now, we know that we need to find students who score greater than 89%

so P(X>89) or P(X>1),

finding value of z= 1 which can be easily found using z score tables.

By looking at table we get the value of z = 0.3413

Subtracting the value of z from 1 we get

Percentage of the students scored higher than 89% = 1-0.3413 = 0.6587 or 65.87%

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17