Question

Anna is at the movie theater and has $35 to spend. She spends $9.50 on a ticket and wants to buy some snacks. Each snack costs $3.50. How many snacks, x, can Anna buy? Inequality that represents this situation: 9.50+3.50x≤35 Drag each number to show if it is a solution to both the inequality and the problem situation, to the inequality only, or if it is not a solution.

Answer 1

Answer:

7 is the maximum snacks Anna can buy.

Step-by-step explanation:

Anna is at the movie theater and has $35 to spend.

She spends $9.50 on a ticket. Money left now = $35-9.50=25.50$

She wants to buy some snacks. Let the snacks count be x

Each snack costs $3.50. So, can buy 3.50x snacks

Now given inequality that represents this situation: $9.50+3.50x \leq 35$

Solving this inequality we can find the number of snacks Anna can buy

$3.50x\leq 25.50$

$x\leq 7.2$

Rounding it to whole number we get 7.

So, Anna can buy a maximum of 7 snacks.

We can check the equation:

$9.50+3.50(7)\leq 35$

$9.50+24.50\leq 35$

$34\leq 35$

Hence, 7 is the maximum snacks Anna can buy.

Answer 2

Answer with explanation:

Amount of money Possessed by Anna = $ 35

Money spent on ticket = $9.50

Money spent on Snacks = $ 3.50

Let x number of snacks, which will be least number of snacks that Anna can buy.

transforming the situation in terms of inequality

→9.50 +3.50 x≤ 35

→9.50 -9.50+3.50 x≤35-9.50

→3.50 x≤25.50

Dividing both sides by 3.50, we get

→x≤7.3(approx)

which can't be number of Snacks, as it will be an integral value.

So, minimum number of snacks with given amount of money = 7

So, Anna can buy snacks(x)={x:x≤7,x=1,2,3,4,5,6,7}=At most 7.