2. The heights of horses who compete in the Melbourne Derby are

normally distributed with a mean of 66 inches and a standard

deviation of 3 inches.

(a) If a horse is randomly selected from this race, what height

would you expect it to be?

(b) What percentage of horses who compete in the Melbourne

Derby have heights between 60 and 72 inches?

(c) In a particular year there are 25 horses running in the

Melbourne Derby. How many of these would you expect

to have a height of less than 63 inches?

66 inches ;

95.44 ;

4

Step-by-step explanation:

Given that:

Normal distribution with;

Mean(m) = 66 inches

Standard deviation (s) = 3 inches

A.) If a horse is chosen at random, Expected height will be equal to the mean of the distribution = 66 inches

B.) percentage horses who are between 60 and 72 inches

Zscore = ( x - m) / s

x = 60

Zscore = (60 - 66) / 3

Zscore = - 6/3

= - 2

P(Z < - 2) = 0.0228

Zscore = ( x - m) / s

x = 72

Zscore = (72 - 66) / 3

Zscore = 6/3

= 2

P(Z < 2). = 0.9772

Between 60 and 72

0.9772 - 0.0228 = 95.44

c) In a particular year there are 25 horses running in the Melbourne Derby. How many of these would you expect to have a height of less than 63 inches?

Zscore = ( x - m) / s

x = 63

Zscore = (63 - 66) / 3

Zscore = - 3/3

= - 1

P(Z < - 1) = 0.1587

0.1587 * 25 = 3. 967 = 4

Area =  .

Step-by-step explanation:

Given : Vanessa uses the expressions (3x² + 5x + 10) and (x² – 3x – 1) to represent the length and width of her patio.

To find : Which expression represents the area (lw) of Vanessa’s patio?

Solution : We have given

Length =  (3x² + 5x + 10)

Width =  (x² – 3x – 1)

Area=  length *  width .

Area = (3x² + 5x + 10) *  (x² – 3x – 1).

Area = 3x²  (x² – 3x – 1) + 5x  (x² – 3x – 1) + 10  (x² – 3x – 1).

Area = .

Combine like terms

Area =  .

Area =  .

Area =  .

Therefore, Area =  .