Question

I need help

Answer 1

With this

Answer 2

What is the sum of the student's four test scores of 75, 83, 89, and 75?

@Atrineas

Answer 3

322 @Directrix

Answer 4

What is it that your trying to do?

Answer 5

\[\frac{ 75 + 83 + 89 + 75 + t_5 }{ 5 } = 80\]

Answer 6

i followed this:
\[\frac{ t_1 + t_2 + t_3 ... + t_n }{ \text{number of tests taken} } = \text{average}\]

t_1, t_2, t_3 ... t_n all represent the test scores.

Answer 7

... Ummm so what do I do?

Answer 8

I'm thinking it's suppose to look like this
\[\frac{ 75_{1} 83_{2} 89_{3} 75_{4} }{ Number :of: test: taken} = average \]
(Sorry for the dots but didn't know why to make it more readable then this)

Answer 9

I could be very wrong too

Answer 10

soo... the current average is 64.4?

Answer 11

I'm not sure

Answer 12

or 80.5?

Answer 13

But even if that is the current average I still don't know how to get the next number that will get you an exact average of 80.... >~<

Answer 14

@agent0smith can you help please... >~<

Answer 15

Sorry @Atrineas I'm just as stuck as you are

Answer 16

It's ok @ScarlettFarra2000

Answer 17

@calculusxy already gave you what you need.

Answer 18

really?

Answer 19

So I just need to find the value of T in her solution right...?

Answer 20

Yes

Answer 21

Ok I'll try to do that thanks

Answer 22

Eh how do I find t srry for asking

Answer 23

(322 + x)/5 = 80 where x is the fifth score

Solve for x:

322 + x = 400 @Atrineas

Answer 24

so x=78...?

Answer 25

@Directrix

Answer 26

Let's check.

Add 322 which is the sum of the first four tests to 78 and divide by 5. If you get 80, then 78 is the correct test score to give a minimum average of 80.

(Also, we are assuming that the tests are equally weighted when determining the average.)

322 + 78
--------- = ? @Atrineas Is that 80?
5

Answer 27

Yes it does

Answer 28

>>so x=78...?

Correct.

Answer 29

Yeah but when I went to put it in mathxl the numbers changed...
@Directrix

Answer 30

I don't know what this is: mathxl

What numbers were changed?

Answer 31

This