IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What percentage of people have an
IQ score between 82 and 98, to the nearest tenth?

33.2%

Step-by-step explanation:

Calculate the z-scores.

z = (x − μ) / σ

z₁ = (82 − 100) / 15

z₁ = -1.2

z₂ = (98 − 100) / 15

z₂ = -0.133

Use a chart or calculate to find the probability.

P(-1.2 < Z < -0.133) = P(Z < -0.133) − P(Z < -1.2)

P(-1.2 < Z < -0.133) = 0.4470 − 0.1151

P(-1.2 < Z < -0.133) = 0.3319

Rounded to the nearest tenth, the percentage is 33.2%.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>