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SSSSS [86.1K]
3 years ago
9

Ms. Washington has 18 students in her class. She wants to send 3 of her students to pick up books for the class. How many combin

ations of students can she choose? Question 10 options:
Mathematics
1 answer:
zimovet [89]3 years ago
4 0
This is a concept of combination, the question requires us to find out how many combinations of students a teacher can choose if she sends 3 of her student to pick up books for the class. Since there is no specific order in selection, the combination will be given by:
nCr
n=Total number of the sample
r=number of students o be selected
thus the combination will be:
18C3
=816
This implies that there are 816 ways in which the Ms Washington can select the students


 
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Answer:

The answer is below

Step-by-step explanation:

A) i)

For Anna initially, she has $0 from making 0 envelopes. After making 400 envelopes she has $20. 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)

For Jason initially, he 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-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-0=\frac{10-0}{250-0}(x-0)\\\\y=\frac{1}{25} x

The table is:

x:   200     400       600     800     1000

y:    8         16           24        32       40

The graph is plotted using geogebra online graphing

b) From the table above we can see that Anna makes more stuffing than Jason.

c) Anna has a savings of $100. 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

We can see from the graph that there is a y intercept at 100. That is the earnings starts from 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)

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

D) The line pass through (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}

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