the Children's Hospital sponsors a 10k race to raise money. it receives $75 per race entry and $10,000 in donations but it must spend $25 per race entry to cover the cost of the race.write and solve an inequality to determine the number of race entries the charity needs to raise at least $100,000.

Respuesta :

Answer:

There would need to be at least 1,800 entries in order for them to meet their goal.

Step-by-step explanation:

To find this, we must write an equation that helps solve it. We know that for every entry (x), we gain 75 and lose 25. We can express this as the following.

75x - 25x

Then we can add the number of donations.

75x - 25x + 10,000

Then we can set it as greater than or equal to the goal number.

75x - 25x + 10,000 ≥ 100,000

Now we solve this using the order of operations.

75x - 25x + 10,000 ≥ 100,000 ----> Combine like terms

50x + 10,000 ≥ 100,000 ------> Subtract 10,000

50x ≥ 90,000 -----> Divide by 50

x ≥ 1,800

Answer:

[tex]x\geq 1800[/tex]

Step-by-step explanation:

Given :

Children's Hospital sponsors a 10k race to raise money.

It receives $75 per race entry .

It receives  $10,000 in donations

It must spend $25 per race entry to cover the cost of the race.

To Find : Write and solve an inequality to determine the number of race entries the charity needs to raise at least $100,000.

Solution :

Let number of race entries be x

We are given that cost of each entry is $75 and it spend $25 of each entry to cover the cost of race .

So, they are remaining with $75-$25=$50 per race entry

So, the actual amount they are getting from each entry = $50

Since there are x entries so cost of x entries will be $50x

And they are getting $10,000 in donations .

So, Total amount they are getting = $50x+$10,000

Since the charity needs to raise at least $100,000.

So, inequality becomes :

⇒[tex]50x+10,000\geq 100,000[/tex]

⇒[tex]50x\geq 100,000-10,000[/tex]

⇒[tex]50x\geq 90,000[/tex]

⇒[tex]x\geq \frac{90,000}{50}[/tex]

[tex]x\geq 1800[/tex]

Thus number of race entries should be at least 1800 .