anonymous
  • anonymous
Jim has received scores of 74and 80 on his first two 100 point tests. What score must he get on his third 100 point test to keep an average of 83 or greater? Third test >= 'blank' (If the average is greater then the possible number of points, enter DNE)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ok, so (74 + 80 + x)/3 = 83 - does that make sense?
myininaya
  • myininaya
inequality
anonymous
  • anonymous
So do I divide the three then?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
You're right myin, but since we know the limit, 100 - aren't we just finding the minimum number that will give you an 83 average? Or am I reading/doing this wrong?
anonymous
  • anonymous
Lulu, you multiply both sides of the equation by 3
anonymous
  • anonymous
(Which gets rid of the /3 portion on the left)
anonymous
  • anonymous
the answer doesn't have to be 83, it can be a higher average
myininaya
  • myininaya
we want the average to be greater than or equal to 83 so we have \[\frac{74+80+x}{3} \text { is the average } \] we want \[\frac{74+80+x}{3} \text{ we want this to be greater than or equal to 83} \] symbolically this looks like \[\frac{74+80+x}{3} \ge 83 \]
myininaya
  • myininaya
but once you get x=100 you could have also wrote x>=100 in the end so your way would have worked fine i just like my equations/inequalities read like my sentences
myininaya
  • myininaya
that was an example by the way with the x=100
myininaya
  • myininaya
i didn't solve this sorry lol
anonymous
  • anonymous
So, I have a question that might be a bit silly. How do I work with the division of the equation. For the life of me I can't remember how to work an equation like that.
anonymous
  • anonymous
Hey lulu - sorry I stepped away for a bit - so your question is regarding how to deal with the division of the equation?
anonymous
  • anonymous
Yes, I'm not sure how to do it. Haven't covered it in a while and I can't remember how to go about it.
anonymous
  • anonymous
No problem! So, the rule of thumb is when you multiply/divide, you need to do the same thing to both sides of the equation.
anonymous
  • anonymous
So, in our case, you want to multiply both sides of the equation by 3, like this: \[(74 + 80 + x)/3 *3 = 83 * 3\]
anonymous
  • anonymous
Oooh I get it, thank you so much!
anonymous
  • anonymous
Awesome - thanks for sticking with me, I got confused myself there for a bit :P. And sorry for going quiet for a bit there - great work lulu, thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.