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ELEN [110]
3 years ago
11

Matt had 60 questions correct on a Percent’s Chapter Test that had 150 one-mark questions. What was his mark written as a percen

tage?

Mathematics
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

His mark, as a percentage, was of 40%.

Step-by-step explanation:

Mark as a percentage:

His mark as a percentage is the number of questions correct multiplied by 100% and divided by the number of questions.

In this problem:

60 questions correct out of 150. So

60*100%/150 = 40%

His mark, as a percentage, was of 40%.

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For each of the following, state the equation of a perpendicular line that passes through (0, 0). Then using the slope of the ne
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Answer:

The answer is below

Step-by-step explanation:

a) y=3x-1

The standard equation of a line is given by:

y = mx + c

Where m is the slope of the line and c is the intercept on the y axis.

Given that y=3x-1, comparing with the standard equation of a line, the slope (m) = 3, Two lines with slope a and b are perpendicular if the product of their slope is -1 i.e. ab = -1. Let the line perpendicular to y=3x-1 be d, to get the slope of the perpendicular line, we use:

3 × d = -1

d = -1/3

To find the equation of the perpendicular line passing through (0,0), we use:

y-y_1=d(x-x_1)\\d\ is\ the \ slope:\\y-0=-\frac{1}{3} (x-0)\\y=-\frac{1}{3}x

To find  x if the point P(x, 4) lies on the new line, insert y = 4 and find x:

y=-\frac{1}{3}x\\ 4=-\frac{1}{3}x\\-x=12\\x=-12

b) y=1/4 x+2

Given that y=1/4 x+2, comparing with the standard equation of a line, the slope (m) = 1/4. Let the line perpendicular to y=1/4 x+2 be f, to get the slope of the perpendicular line, we use:

1/4 × f = -1

f = -4

To find the equation of the perpendicular line passing through (0,0), we use:

y-y_1=f(x-x_1)\\f\ is\ the \ slope:\\y-0=-4 (x-0)\\y=-4x

To find  x if the point P(x, 4) lies on the new line, insert y = 4 and find x:

y=-4}x\\ 4=-4x\\x=-1

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Let the numbers be x and y ∴ 4x -2y =-16......................Eqn 1

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