All students who complete math iii course at little hills high school take a common final exam. the exam scores are normally distributed with a mean of 105 and a standard deviation of 16. a. kyle and ethan are algebra 2 students who took the final exam. kyle's score was 135 and ethan's score was 93. calculate the z-score for each student. round your answers to the nearest tenth. b. what percent of the students had a final exam score lower than ethan's score?

a)

Kyle's z-score was 1.9 to the nearest tenth

Ethan's z-score was -0.8 to the nearest tenth

b)

The percent of the students had a final exam score lower than Ethan's score was 21.19%

Step-by-step explanation:

a) Lets revise how to find the z-score

- The rule the z-score is z = (x - μ)/σ , where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- The exam scores are normally distributed with a mean of 105 and a

  standard deviation of 16

∴ μ = 105 and σ = 16

- Kyle and Ethan are took the final exam

- Kyle's  score was 135

- Ethan's score was 93

- Lets find the z-score for each one

∵ Kyle's  score was 135

∴ x = 135

∵ μ = 105 and σ = 16

∵ z-score = (x - μ)/σ

∴ z-score for Kyle = (135 - 105)/16 = 30/16 = 15/8 = 1.875

* Kyle's z-score is 1.9 to the nearest tenth

∵ Ethan's  score was 93

∴ x = 93

∵ μ = 105 and σ = 16

∵ z-score = (x - μ)/σ

∴ z-score for Ethan = (93 - 105)/16 = -12/16 = -3/4 = -0.75

* Ethan's z-score is -0.8 to the nearest tenth

b) To find the percent of students with a lower exam score than Ethan

   you will asking to find the proportion of area under the standard

   normal distribution curve for all z-scores < -0.8

- It can be read from a z-score table by referencing a z-score of -0.8

- Look to the attached file

∴ The value from the table is 0.2119

- To change it to percent multiply it by 100%

∴ 0.2119 × 100% = 21.19%

* The percent of the students had a final exam score lower than

  Ethan's score was 21.19%

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°