Question

1) Anna and Jason have summer jobs stuffing envelopes for two different companies. Anna earns $20 for every 400 envelopes she finishes. Jason earns $10 for every 250 envelopes he finishes.
A) Create tables, draw graphs, and write equations that show the earnings, y, as functions of the number of envelopes stuffed, x, for both Anna and Jason.

B) Who makes more from stuffing the same number of envelopes? Explain how you know using the information above.

C) Anna has savings of $100 at the beginning of the summer and she saves all of her earnings from her job. Graph her savings as an equationof the number of envelopes she’s stuffed, n. How does this graph compare to her previous earnings graph? What is the meaning of the initial value and slope in each case?

D)You survey 6 other employees and findthe following data on envelopesstuffed (x) and dollars earned(y). Plot a trend lineon the graph. Find the equation of the trend line.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

A) i)

For Anna initially, she has $0 from making 0 envelopes. Anna earns $20 for every 400 envelopes. Let x represent the number of envelopes and y the earnings. Hence this can be represented by the points (0, 0) and (400, 20). Using the equation of a line:

$y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-0=\frac{20-0}{400-0}(x-0)\\\\y=\frac{1}{20} x$

The table is:

x:   200     400       600     800     1000

y:    10        20          30        40       50

ii)

Jason has $0 from making 0 envelopes. For every 250 envelopes he has $10. Let x represent the number of envelopes and y the earnings. Hence this can be represented by the points (0, 0) and (250, 10). Using the equation of a line:

$y-0=\frac{10-0}{250-0}(x-0)\\\\y=\frac{1}{25} x$

x:   200     400       600     800     1000

y:    8         16           24        32       40

The graph is plotted using geogebra online graphing

b) From the table of Anna and Jason we can see that Anna makes more stuffing than Jason for the same number of envelopes.

c) Anna has a savings of $100 initially. Hence this can be represented by the points (0, 100) and (250, 10). Using the equation of a line:

$y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-100=\frac{20-0}{400-100}(x-0)\\\\y=\frac{1}{15} x+100$

Therefore from the graph there is a y intercept at 100. That is the earnings starts initially is $100.

The equation of a line is given as y = mx + b, where m is the slope and b is the y intercept (initial value of y at x = 0)

For the first graph, the slope is 1/20 with an initial value of 0 while for the second graph the slope is 1/15 with an initial value of 100

D) The trend line pass through the points (10, 10) and (100, 40), hence:

$y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-10=\frac{40-10}{100-10}(x-10)\\\\y-10=\frac{1}{3} (x-10)\\\\y=\frac{1}{3}x+\frac{20}{3}$