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Serhud [2]
3 years ago
6

Use the graph representing bacteria decay to estimate the domain of the function and solve for the average rate of change across

the domain.

Mathematics
2 answers:
nasty-shy [4]3 years ago
5 0

Answer:

Domain is [0,18]

Average rate of change of bacteria per minute is -3333.33.

Step-by-step explanation:

Domain is the possible values of x.

From the graph, it's clear that the x values ranges from 0 to 18 as the end points of the graph are (0,60) and (18,0)

So, domain of the function is [0,18].

Now, average rate of change is negative as there is bacteria decay.

Average rate of change is given as the negative ratio of y intercept to the x intercept.

Here, x-intercept is 18 and y-intercept is 60000.

So, average rate of change is given as,

R_{avg}=-\frac{60000}{18}=-3333.33

Therefore, average rate of change of bacteria per minute is -3333.33.

ollegr [7]3 years ago
3 0
  • 0 ≤ x ≤ 18
  • -3.33
<h3>Further explanation</h3>

The domain of a function is the set of values of the independent variable for which a function is defined. The domain is located along the x-axis where the graph is defined from the starting point to the endpoint.

From the graph, we can see that:

  • the horizontal axis is the amount of time (in minutes)
  • the vertical axis is the number of bacteria (in thousands)
  • (0, 60) is a y-intercept
  • (18, 0) is an x-intercept

Also from the graph, we can observe that the values ​​of x start from x = 0 to x = 18.

Thus the domain of the function is \boxed{ \ [0, 18] \ or \ 0 \leq x \leq 18 \ }

The formula for the average rate of change on a given interval [a, b] is as follows:

\boxed{ \ \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a} \ }

Because the domain of the function is [0, 18], we prepare:

  • \boxed{(0, 60) \ as \ (a, f(a))} and
  • \boxed{(18, 0) \ as \ (b, f(b))}

And now, let us solve for the average rate of change across the domain.

\boxed{ \ \frac{\Delta y}{\Delta x} = \frac{0 - 60}{18 - 0} \ }

\boxed{ \ \frac{\Delta y}{\Delta x} = \frac{- 60}{18} \ }

\boxed{\boxed{ \ \frac{\Delta y}{\Delta x} = -3.33 \ }}

Note that the y-values change down 3.33 units in thousands every time the x-values change 1 unit in minutes, at intervals of the domain.

We conclude that from t = 0 to t = 18 (in minutes), every 1 minute as many as 3.33 thousand bacteria decay.

<h3>Learn more</h3>
  1. What are the domain and range of the function f(x) = 3x + 5? brainly.com/question/3412497
  2. What is the range of the given function? {(–2, 0), (–4, –3), (2, –9), (0, 5), (–5, 7)} brainly.com/question/1435353
  3. The piecewise-defined functions brainly.com/question/9590016

Keywords: use, the graph, representing, bacteria decay, to estimate, the domain of the function, and, solve, the average rate of change, axis, horizontal, vertical, intervals, minutes, thousand, 0, 18, 60, -3.33

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Step-by-step explanation:

Step 1

We arrange the given the following data from lowest to the highest

32.0, 32.1, 32.1, 32.2, 32.4, 32.5, 32.5, 32.5, 32.8, 32.8, 32.8, 32.8, 32.9, 32.9, 33.0, 33.0, 33.2, 33.2, 33.2, 33.2, 33.5, 33.6, 33.8, 33.8, 33.9, 34.0, 34.0, 34.1, 34.1, 34.3, 34.4, 34.5, 34.6, 34.6, 34.6, 34.7, 34.7, 35.0, 35.0, 35.1, 35.1, 35.2, 35.4, 35.5, 35.5, 35.5, 35.5, 35.6, 35.6, 36.1, 36.3, 36.4, 36.4, 36.4, 36.5, 36.6, 36.7, 36.7, 37.0, 37.1, 37.4, 37.4, 37.6, 37.8

a. Construct a stem-and-leaf display for the data.

Stem and leaf Display

Stem | Leaf

32 | 0,1,1,2,4,5,5,5,8,8,8,8,9,9

33 | 0,0,2,2,2,2,5,6,8,8,9

34 | 0,0,1,1,3,4,5,6,6,6,7,7

35 | 0,0,1,1,2,4,5,5,5,5,6,6,

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b. Calculate the median and quartiles of these data.

Number of terms = 64

1) Median = 1/2(n + 1)th value

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= 1/2(64 + 1)th

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This means it is between the 32nd and 33rd value

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33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

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2) First Quartile

1/4(n + 1)th value

n = 64

= 1/4(64 + 1)th

= 1/4(65)th

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This means it is between the 16th and 17th value

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17tj value = 33.2

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Q1 --> 33.1

3)Second Quartile = Median

1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

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= 69.1/2

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Q2 --> 34.55

Third Quartile

3/4(n + 1)th value

n = 64

= 3/4(64 + 1)th

= 3/4(65)th

= 48.75 th value

This means it is towards the 49th value

32nd = 34.5

Hence,

Q3 --> 35.6

Inter Quartile range

Q3 - Q1

= 35.6 - 33.1

= 2.5

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