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svp [43]
4 years ago
8

If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes

t for a 90 average?
Mathematics
2 answers:
gladu [14]4 years ago
8 0
Average = sum/count

current count = 4
current sum = 76+92+89+97=354

we need average = 90 = sum/5
so, we need sum = 90*5=450

To get sum 450, he needs to score 450-354=96 on his best test.

Now, (76+92+89+97+96)/5=90
Keith_Richards [23]4 years ago
4 0
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
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x = additional trips = 4

We then put the value of x into the equation

= 3400 + 1200x

= 3400 + 1200(4)

= 3400 + 4800

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Read 2 more answers
Check all of the ordered pairs that satisfy the equation below.y=2/5x
zvonat [6]

We are going to do this step by step:

Let's start with option A

A.

X = 25 and Y = 5/2

y=\frac{2}{5}\cdot x=\frac{2}{5}\cdot25=\frac{2\cdot25}{5}=\frac{50}{5}=\frac{5\cdot10}{5\cdot1}=10

In this case, when X = 25 , then Y = 10 ,which is different to 5/2

B.

X = 14 , Y = 35

y=\frac{2}{5}\cdot14=\frac{28}{5}

In case, when X = 14, then Y = 28/5, which is different to 35

C.

X = 40 , Y = 24

y=\frac{2}{5}\cdot40=\frac{80}{5}=16

Similarly, in this case, when X = 40, then Y = 16, which is different to 24

D.

X = 10 , Y=4

y=\frac{2}{5}\cdot10=\frac{20}{5}=4

Now, in this case, we can see that when X = 10, then Y = 4 which is the same as the given value of Y

E.

X = 50 , Y = 20

y=\frac{2}{5}\cdot50=\frac{100}{5}=20

In this case, the values of Y are also the same.

F.

X = 30 , Y = 12

y=\frac{2}{5}\cdot30=\frac{60}{5}=12

Again, in this case, the values of Y are the same, so the pair satisfies the equation.

In conclusion: options D, E and F satisfy the equation

3 0
1 year ago
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