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sladkih [1.3K]
3 years ago
6

The line whose equation is 3x - 5y = 4 is dilated by a scale factor of 3

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0

Answer:

Answer 1.

Step-by-step explanation:

The complete question is:

The line whose equation is 3x - 5y = 4 is dilated by a scale factor of 3 centered at the origin. Which statement is correct?

1. The image of the line has the same slope as the pre-image but a different y-intercept.

2. The image of the line has the same y-intercept as the pre-image but a different slope.

3. The image of the line has the same slope and the same y-intercept as the pre-image.

4. The image of the line has a different slope and a different y-intercept from the pre-image.

Recall that a dilation by a scale factor r centered at the origin takes a point (x,y) and maps it to the point (r*x,r*y).

Consider also a line of equation ax+by=c. From this equation, the slope of the line is given by the number \frac{-a}{b} and the y-intercept is given by \frac{c}{b}.

Consider the given equation 3x-5y =4. If we replace (x,y) with (3x,3y), we get

3(3x)-5(3y) = 4 = 3 (3x-5y) = 4

which is equivalent to

3x-5y = \frac{4}{3}

Note that comparing the equations 3x-5y=4 and 3x-5y=\frac{4}{3} the values of a and b are equal but the value of c is different. This means that they have the same slope but a different y-intercept.

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To study the output of a machine that fills boxes with cereal, a quality control engineer weighed 150 boxes of Brand X cereal. T
scoundrel [369]

Answer:

\sigma = 0.07770

Step-by-step explanation:

Given

The above table

Required

The standard deviation

First, calculate the class mid-point. This is the average of the class interval.

The mid-point x is:

x_1 = \frac{1}{2}(15.3 + 15.6) =\frac{1}{2} * 30.9 = 15.45

x_2 = \frac{1}{2}(15.6 + 15.9) =\frac{1}{2} * 31.5 = 15.75

x_3 = \frac{1}{2}(15.9 + 16.2) =\frac{1}{2} * 32.1 = 16.05

x_4 = \frac{1}{2}(16.2 + 16.5) =\frac{1}{2} * 32.7 = 16.35

x_5 = \frac{1}{2}(16.5 + 16.8) =\frac{1}{2} * 33.3 = 16.65

So, we have:

\begin{array}{cccccc}x & {15.45} & {15.75} & {16.05} & {16.35} & {16.65} \ \\ f & {14} & {25} & {84} & {18} & {9} \ \end{array}

Calculate mean

\bar x = \frac{\sum fx}{\sum f}

\bar x = \frac{15.45 * 14 + 15.75 * 25 + 16.05 * 84 + 16.35 * 18 + 16.65 * 9}{14 + 25 + 84 + 18 + 9}

\bar x = \frac{2402.4}{150}

\bar x = 16.016

The standard deviation is:

\sigma = \sqrt{\frac{\sum(x_i - \bar x)^2}{\sum f}}

\sigma = \sqrt{\frac{(15.45 - 16.016)^2+ (15.75  - 16.016)^2+ (16.05  - 16.016)^2+ (16.35   - 16.016)^2+ (16.65  - 16.016)^2}{14 + 25 + 84 + 18 + 9}}

\sigma = \sqrt{\frac{0.90578}{150}}

\sigma = \sqrt{0.00603853333}

\sigma = 0.07770

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goldenfox [79]

Answer: y=23

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