Refer to Exercise 3.145. Use the uniqueness of moment-generating functions to give the distribution of a random variable with moment-generating function m(t) = (.6et + .4)3.

The random variable with m(t) = (0.6e^t+ 0.4)^3

follows binomial random variable with parameters p= 0.6 and n= 3

Step-by-step explanation:

Given that p + q = 1

q = 1 - p

The general form of moment generating function, MGF, m(t) is given as

m(t) = [pe^t + q]^n for a binomial distribution

Comparing this to the moment-generating function to [0.6e^t + 0.4]^3

These m(t) functions are exactly the same with

p= 0.6,

q = 1 - 0.6

q= 0.4

n= 3.

Thus, the random variable with m(t) = (0.6e^t+ 0.4)^3

follows binomial random variable with parameters p= 0.6 and n= 3

i can answer this question sweetie!

step-by-step explanation:

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841