If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89

0.60515

Step-by-step explanation:

Given that:

Population proportion p = 0.9

Sample size = 64

Probability that sample proportion is more Than 0.89

P(p" > 0.89)

(p" - p) ÷ sqrt(p(1 - p) /n)

P(Z > p")

P(Z > (0.89 - 0.9) ÷ sqrt(0.9(1 - 0.9) /64))

P(Z > (-0.01) ÷ sqrt(0.9 * 0.1) / 64))

P(Z > (-0.01) / sqrt(0.00140625))

P(Z > - 0.26667) = 0.6051

Using the Z probability calculator :

P(Z > - 0.26667) = 0.60515

50 min

step-by-step explanation:

they will be decreasing both their times in half

60 / 2 = 30

40/2 = 20

30 + 20 = 50

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2