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Ray Of Light [21]
2 years ago
11

Follow this link to view Jamie’s work. Critique Jamie’s work by explaining the reasonableness of the scenario and the accuracy o

f the two parts of the graph he described . For any portion of Jamie’s response in which he has made an error, provide and explain an alternative response. To enhance Jamie’s explanation, describe both sections of the piecewise function using linear equations. Include any limits to the domain.

Mathematics
1 answer:
Stella [2.4K]2 years ago
7 0

Answer:

See  below

Step-by-step explanation:

This graph could describe a relay race.

Jamie writes a scenario that can be modeled by the piecewise function on the graph.

The graph has time, in minutes on the x-axis and the distance, in miles on the y-axis.

The first part of the graph is the line starting from the origin through to the point (12, 1) then the second part starts from the point (12, 1) and ends on the point (20, 2).

1. A runner begins the race and runs at a steady pace for 1 minute.

  • This is incorrect as describes the wrong axis.

It should say: a runner begins a race and runs at a steady pace for 12 minutes, completing 1 mile distance.

<u>The equation for this part:</u>

  • y = mx + b
  • b = 0 as it starts at origin.
  • m = 1/12
  • y = 1/12x
  • x = [0, 12], y = [0, 1]

2. His partner then takes over and runs at a steady, but faster, pace for another 1 minute.

  • Again, as above, wrong axis is described.

It should say: his partner then takes over and runs at a steady, but faster, pace for another 8 minutes, completing the last 1 mile of the race.

<u>The equation for this part:</u>

  • b =1, as a starting point
  • m = (2-1)/(20-12) = 1/8
  • y = 1/8x + 1
  • x = (12, 20], y = (1, 2]

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--------------------------

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14 = 7c
14/7 = 7c/7
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c = 2
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b + c = 7
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<em>b + (2) = 7
</em><em />Find b by isolating it. subtract 2 from both sides
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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?
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Answer:

(a) Selection = 6724520

(b) At\ most\ 12 = 6553976

(c) At\ most\ 8 = 6066720

(d) At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

Step-by-step explanation:

Given

Colors = 8

Balloons = 28 --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

n => 28 + 8-1= 35

r = 28

Selection = ^{35}^C_{28

Selection = \frac{35!}{(35 - 28)!28!}

Selection = \frac{35!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29}{7!}

Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}

Selection = \frac{33891580800}{5040}

Selection = 6724520

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

n => 15 + 8 -1 = 22

r = 15

Selection of at least 13 =

At\ least\ 13 = ^{22}C_{15}

At\ least\ 13 = \frac{22!}{(22-15)!15!}

At\ least\ 13 = \frac{22!}{7!15!}

At\ least\ 13 = 170544

Ways of selecting at most 12  =

At\ most\ 12 = Total - At\ least\ 13 --- Complement rule

At\ most\ 12 = 6724520- 170544

At\ most\ 12 = 6553976

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

n => 19+ 8 -1 = 26

r = 19

Selection of at least 9 =

At\ least\ 9 = ^{26}C_{19}

At\ least\ 9 = \frac{26!}{(26-19)!19!}

At\ least\ 9 = \frac{26!}{7!19!}

At\ least\ 9 = 657800

Ways of selecting at most 8  =

At\ most\ 8 = Total - At\ least\ 9 --- Complement rule

At\ most\ 8 = 6724520- 657800

At\ most\ 8 = 6066720

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

n =6+6-1 = 11

r = 6

Selection =

^{11}C_6 = \frac{11!}{(11-6)!6!}

^{11}C_6 = \frac{11!}{5!6!}

^{11}C_6 = 462

Using inclusion/exclusion rule of two sets:

Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)

Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462

Only\ 12\ red\ and\ only\ 8\ blue\ = 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  6724520 - 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

3 0
2 years ago
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