1. A painter earns $15 per hour. What is the minimum number of hours he must work to earn at least $200? Write an inequality to represent this situation and solve.

Let x be the minimum number of hours the painter must work to earn at least $200.

The inequality to represent this situation is:
15x ≥ 200.
To solve for x, we divide both sides of the inequality by 15:
x ≥ 200/15.
Simplifying the right side of the inequality gives x ≥ 40/3.
Since we cannot have a fraction of an hour, the minimum number of hours the painter must work to earn at least $200 is 14 hours.

To find the minimum number of hours the painter must work, we can set up an inequality.

Let's assume the minimum number of hours the painter must work to earn at least $200 is represented by the variable "h".

The painter earns $15 per hour, so the total earnings will be the product of the hourly rate and the number of hours worked, which can be represented as 15h.

We want to find the minimum number of hours "h" that satisfies the condition of earning at least $200. Therefore, we can set up the following inequality:

15h ≥ 200

Now let's solve for "h".

Divide both sides of the inequality by 15 to isolate "h":

15h/15 ≥ 200/15

Simplifying:

h ≥ 13.33

Since the painter cannot work a fraction of an hour, the minimum number of hours the painter must work to earn at least $200 is 14 hours.

To solve this problem, we need to set up an inequality expression to represent the situation.

Let's assume that the number of hours the painter must work is x. Since the painter earns $15 per hour, the total amount he earns after working x hours is 15x.

According to the question, the painter needs to earn at least $200. Therefore, we can write the inequality:

15x ≥ 200

To solve for x, we need to isolate the variable on one side of the inequality. In this case, we can divide both sides by 15:

x ≥ 200/15

Simplifying the right side gives:

x ≥ 40/3

So, the minimum number of hours the painter must work to earn at least $200 is 40/3, which is approximately 13.33 hours.