Worksheet titled "Linear Inequalities" with three word problems involving real-life situations requiring the use of linear inequalities to solve.

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Problem 1:

> In 2022–2023, Washington, Colorado, received 278 inches of annual snowfall. From 2020–2021, the city received 256 inches of annual precipitation.
> Write an inequality that expresses how much greater the snowfall was in 2022–2023 than in 2020–2021.

#### Solution:
Let $ x $ represent the amount of snowfall in 2022–2023.
We know:
- Snowfall in 2022–2023 = 278 inches
- Precipitation in 2020–2021 = 256 inches

We want to express how much greater the snowfall was in 2022–2023 compared to 2020–2021.

So,
$$
x > 256
$$

But since we know the exact value (278), we can write:
$$
278 > 256
$$

Alternatively, if we're comparing differences:
$$
\text{Difference} = 278 - 256 = 22 \text{ inches}
$$

So, the snowfall in 2022–2023 was 22 inches more than in 2020–2021.

Inequality:
$$
\text{Snowfall}_{2022–2023} > \text{Precipitation}_{2020–2021}
\Rightarrow 278 > 256
$$

✔ This is a true statement, but if we're writing an inequality for "how much greater," it's simply:
$$
\text{Snowfall} > 256
$$

Or, if asking for the difference:
$$
\text{Snowfall} - 256 > 0 \Rightarrow 278 - 256 = 22 > 0
$$

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Problem 2:

> A professional basketball team needs approximately 100 minutes of practice per day. For the team under the condition that it cannot exceed more than 20 players in total, write the inequality which can relate to this goal.

#### Solution:
Let $ p $ be the number of players.

The condition says: no more than 20 players, so:
$$
p \leq 20
$$

This is the inequality representing the constraint.

If we consider time, and assume each player needs some practice time, but no specific time per player is given, then the key restriction is just the number of players.

So, the inequality is:
$$
p \leq 20
$$

✔ Final answer: $ p \leq 20 $

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Problem 3:

> A car company produces two types of cars: A and B, each of which requires processing in two machines. The first machine can be used at most 20 hours. The second machine can be used at most 30 hours. Car A requires 2 hours on machine one and 1 hour on machine two. Car B requires 1 hour on machine one and 3 hours on machine two.
> Suppose the plant wants to maximize profit. Write inequalities for each machine.

#### Solution:
Let:
- $ x $ = number of Car A produced
- $ y $ = number of Car B produced

Machine 1 (max 20 hours):
- Car A uses 2 hours → $ 2x $
- Car B uses 1 hour → $ 1y $
- Total time: $ 2x + y \leq 20 $

Machine 2 (max 30 hours):
- Car A uses 1 hour → $ 1x $
- Car B uses 3 hours → $ 3y $
- Total time: $ x + 3y \leq 30 $

Also, non-negativity constraints:
$$
x \geq 0, \quad y \geq 0
$$

✔ So, the system of inequalities is:
$$
\begin{cases}
2x + y \leq 20 \\
x + 3y \leq 30 \\
x \geq 0 \\
y \geq 0
\end{cases}
$$

These define the feasible region for production.

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Problem 4 (partial):

> A student wants to buy a gift for his father. He has $50. The gift costs $20. He also wants to buy a card that costs $5. How many gifts can he buy?

Wait — actually, let me re-read this carefully.

From the visible text:
> A student wants to buy a gift for his father. He has $50. The gift costs $20. He also wants to buy a card that costs $5. How many gifts can he buy?

Let’s suppose he wants to buy g gifts and one card.

Cost of gifts: $ 20g $
Cost of card: $ 5 $
Total cost: $ 20g + 5 \leq 50 $

Solve:
$$
20g + 5 \leq 50 \\
20g \leq 45 \\
g \leq 2.25
$$

Since $ g $ must be a whole number (can't buy a fraction of a gift),
$$
g \leq 2
$$

✔ So, he can buy at most 2 gifts.

Inequality: $ 20g + 5 \leq 50 $

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Problem 5 (partially visible):

> Use a strategy to determine for a large family gathering. His budget for the gathering is $1000. Each adult costs $25, each child costs $15. Write this as an inequality, and then graph it.

Let:
- $ a $ = number of adults
- $ c $ = number of children

Total cost:
$$
25a + 15c \leq 1000
$$

This is the inequality.

To graph it:
- Intercepts:
- If $ a = 0 $: $ 15c = 1000 \Rightarrow c = \frac{1000}{15} \approx 66.67 $
- If $ c = 0 $: $ 25a = 1000 \Rightarrow a = 40 $

Plot the line $ 25a + 15c = 1000 $, shade below it (since ≤), with $ a \geq 0, c \geq 0 $.

✔ Inequality: $ 25a + 15c \leq 1000 $

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Summary of Solutions:

| Problem | Answer |
|--------|--------|
| 1 | $ 278 > 256 $ or $ \text{Snowfall} > 256 $ |
| 2 | $ p \leq 20 $ |
| 3 | $ 2x + y \leq 20 $, $ x + 3y \leq 30 $, $ x \geq 0 $, $ y \geq 0 $ |
| 4 | $ 20g + 5 \leq 50 $; maximum $ g = 2 $ |
| 5 | $ 25a + 15c \leq 1000 $ |

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If you can provide a clearer version of the image or specify which problem you'd like explained in detail (e.g., graphing, solving, interpreting), I can go deeper into any one of them!