Over the course of one day, a store owner determined that 80% of customers bought a drink and 30% of customers bought a snack. if the store sold 120 snacks that day, how many more drinks than snacks did the store sell?

200

Step-by-step explanation:

Let the total number of customers that day be "x"

30% of total customers bought a snack and that is 120.

So we can say:

30% of total (x) is 120

Note: 30% converted to decimal is 30/100 = 0.3

This can be translated to an algebraic equation as:

0.3x = 120

Now, we can solve for x, the total customers:

0.3x = 120

x = 120/0.3

x = 400

Since, 80% bought drink, we find 80% of 400:

80% = 80/100 = 0.8

0.8 * 400 = 320

The store sold 320 drinks and 120 snacks. So drinks outweigh snacks be:

320 - 120 = 200

So,

200 more drinks than snacks

200 more drinks than snacks did the store sell.

Given,

The percentage of customers who bought a drink is 80%.

The percentage of customers who bought a snack is 30%.

There are 120 snacks sold for the day.

We need to determine that how many more drinks than snacks did the store sell.

Let us assume that the total number of customers be x.

So, the 30% of total customers is 120.

Thus,

Therefore total customers are 400.

The customers who can buy drinks can be calculated as:

Thus,

Hence, 200 more drinks than snacks did the store sell.

To know more about the percentage, please refer to the link:

link

i believe that the answer is

b. y-1=-2(x-4)

hope this !

no

step-by-step explanation: