"If the average cholesterol level is 194 with a standard deviation of 15, what percentage of children have a cholesterol level lower than 199? Answers are rounded to the nearest whole percent."

63%

Step-by-step explanation:

Given the following :

Average cholesterol level or mean (m) = 194

Standard deviation (sd) = 15

what percentage of children have a cholesterol level lower than 199?

P(X < 199)

Assume a normal distribution :

Find the z-score :

Z = (score - mean) / standard deviation

Score = 199

Z = (199 - 194) / 15

Z = 5 / 15

Z - score = 0.3333

P(Z < 0.33) :

Using the z - table ; 0.33 = 0.6293

P(Z < 0.33) = 0.6293

0.6293 * 100% = 62.93%

= 63% (nearest whole percent)

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15