67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are randomly selected, find the probability that a. Exactly 29 of them are spayed or neutered. b. At most 33 of them are spayed or neutered. c. At least 30 of them are spayed or neutered. d. Between 28 and 33 (including 28 and 33) of them are spayed or neutered. PL I NEED THE ANSWER FOR ALL OF THEM

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p  = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

where

a) Probability that exactly 29 of them are spayed or neutered

b) Probability that at most 33 of them are spayed or neutered

c) Probability that at least 30 of them are spayed or neutered

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.