A retail store sells two types of shoes, sneakers and sandals. The store owner pays $8 for the sneakers and $14 for the sandals. The sneakers can be sold for $10 and the sandals can be sold for $17. The owner of the store estimates that she won't sell more than 200 shoes each month, and doesn't plan to invest more that $2,000 on inventory of the shoes. How many of each type of shoe should be stocked in order to maximize her total monthly profit?

134 sneakers and 66 sandals

Step-by-step explanation:

Sneakers= x, sandals= y

Profit rate:

Sneakers are more profitable, so it should be maximized.

Cost:

8x+14y≤2000

x+y≤200

if we assume the maximum of 200 shoes stocked, then:

x= 200-y

8(200-y)+14y ≤ 2000

1600 - 8y +14y ≤ 2000

6y≤400

y≤66 and x=134 is the best option

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5