According to a recent survey, 73% of all customers will return to the same grocery store. Suppose 14 customers are selected at random, what is the probability that: (Round probabilities to 4 decimal places and the expected customers to 2 decimal places)

This question is incomplete, the complete question is;

According to a recent survey, 73% of all customers will return to the same grocery store. Suppose 14 customers are selected at random, what is the probability that: (Round probabilities to 4 decimal places and the expected customers to 2 decimal places)

a) All 14 will return?

b) How many customers would be expected to return to the store ?

a) the probability that All 14 will return is 0.0122

b) The number of customers that would be expected to return to the store is 13.27

Step-by-step explanation:

Given the data in the question;

p = 73% = 0.73

sample size n = 14

a)  the probability that All 14 will return;

P( x = 14 ) = ⁿCₓ × Pˣ ×

we substitute

P( x = 14 ) = ¹⁴C₁₄ × (0.73)¹⁴ ×

P( x = 14 ) = [ 14! / ( 14! × ( 14 - 14)!) ] ×  (0.73)¹⁴ ×

P( x = 14 ) = 1 × 0.0122045 ×

P( x = 14 ) = 1 × 0.0122045 × 1

P( x = 14 ) = 0.0122045 ≈ 0.0122 { 4 decimal place }

Therefore, the probability that All 14 will return is 0.0122

b) Expected Customers;

E(x) = n × p

we substitute

E(x) = 14 × 0.73

E(x) =  13.27  { 2 decimal place }

Therefore, The number of customers that would be expected to return to the store is 13.27

b. 2 and 4

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)