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nasty-shy [4]
3 years ago
5

Ryan spent $26.32 on 8 gallons of gas. How much would he spend for 20 gallons?

Mathematics
2 answers:
zheka24 [161]3 years ago
5 0
He would spend $65.8. 
($26.32 x 20 =526.4)
(526.4 divided by 8 = 65.8
$65.8 would be your answer. 
nalin [4]3 years ago
4 0
$65.80

$26.32/8=$3.29 for a gallon
$3.29 x 20 = $65.80.

Hope this helps
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Suppose you received a score of 95 out of 100 on exam 1 . The mean was 79 and the standard deviation was 8 . If your score on ex
Schach [20]

Answer:

You did the same on both exams.

Step-by-step explanation:

To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:

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Where X = score obtained

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           σ = standard deviation

For Exam 1:

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For Exam 2:

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3 years ago
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

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Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

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Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

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