The weights of Siamese cats are normally distributed with a mean of 6.4 pounds and a standard deviation of 0.8 pounds. If a breeder of Siamese cats has 128 in his care, how many can he expect to have weights between 5.2 and 7.6 pounds?

(1) 106 (3) 98

(2) 49 (4) 111

(4) 111

Step-by-step explanation:

Given that:

mean (μ) = 6.4 pounds

Standard deviation (σ) = 0.8 pounds

Number of Siamese cats (n) = 128

The z score (z) is given by the equation:

, Where x is the raw score.

For a raw score of 5.2 pounds:

For a raw score of 7.6 pounds:

From the normal probability table:

P(5.2 < x < 7.6) = P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668= 0.8664 = 86.64%

Therefore 86.64% of Siamese cats have weights between 5.2 and 7.6 pounds.

The number of Siamese cats that have weights between 5.2 and 7.6 pounds = 86.64% × n = 0.8664 × 128 ≈ 111

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

answer: 20

step-by-step explanation: