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2 years ago

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A) Determine the functions whose graphs are parallel to the graph of the function y=0.5x+10 and intersect the graph of the funct

ion y= -1.5x. The functions are expressed by the formulas.
y = -1.5x + 6, y = 0.5x-6

I WILL AWARD 75 POINTS !

B) Determine the functions whose graphs are parallel to the graph of the function y=0.5x+10 and intersect the graph of the function y= -1.5x. The functions are expressed by the formulas.

y = 0.5x+4, y = 0.5x, y = 3+1.5x

Please answer somebody. Someone help

1 answer:

Sonbull [250]2 years ago

3 0

We want to find parallel lines to the given lines, such that also meet some given restrictions.

The solutions are:

A) y = 0.5*x - 6

B) two solutions:

y = 0.5*x + 4

y = 0.5*x

Before start answering, a general line is given by:

y = a*x + b

Where a is the slope and b is the y-intercept.

Another line will be parallel this if and only if it has the same slope and a different y-intercept.

A) We want to find a line that is parallel to y = 0.5*x + 10 and that intersects y = -1.5*x

So we want another line with a slope of 0.5, then the correct option is:

y = 0.5*x - 6

How do we know that this intercepts the other line?

Because if two lines have different slopes, then the lines are not parallel, meaning that at some given point the lines intercept.

B) Similarly, now we want a line parallel to:

y = 0.5*x + 10

And that intersects:

y = -1.5*x

The correct options are the ones with slope equal to 0.5, then we have two correct options:

y = 0.5*x + 4

y = 0.5*x

If you want to learn more, you can read:

brainly.com/question/16701300

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The shortest way to solve this problem would be:

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Or you could do the longer way, which would be:

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Step 2: Find the surface area of the two ends of the cylinder

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Confidence interval : $21.506$ to $24.493$

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

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So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

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Significance level = α = 0.001

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Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$

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Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table $t_{\frac{\alpha}{2}$ = 3.883

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Substitute the values in the formula

Confidence interval : $23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $23 -3.883 \times \frac{1.72}{\sqrt{20}}$ to $23 + 3.883 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $21.506$ to $24.493$

Hence Confidence interval : $21.506$ to $24.493$

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B={(2,2),(4,2),(6,2),(2,4),(2,6),(1,3),(1,5),(1,1),(3,1),(3,3),(3,5),(4,4),(4,6),(5,1),(5,3),(5,5),(6,4),(6,6)}

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A∩B:{(2,2),(4,2),(6,2),(2,4),(2,6),(4,4),(4,6),(6,4),(6,6)}

A∪B={(1,2),(2,2),(4,2),(6,2),(2,4),(2,6),(4,4),(4,6),(6,4),(6,6),(3,2),(5,2),(1,4),(3,4),(5,4),(1,6),(3,6),(5,6),(1,3),(1,5),(1,1)(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)}

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Step-by-step explanation:

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