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enyata [817]
3 years ago
8

Describe a situation that models a linear pattern and then describe a situation that models a nonlinear pattern. Do not state wh

ich situation is linear and which is nonlinear. Your classmates will need to determine which is which and then write a function that models the linear situation. Your classmates will also need to write 1 ordered pair that is a solution to the linear function and explain its meaning in the context of the situation.

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
6 0

Answer:

When we talk about linear patterns, we refer to those events where events are related with a linear proportion, that is, if one magnitude increases, the other one will to, or if one magnitude increases the other would decrease, and this behaviour with a linear function which is represented as a line.

So, one example of linear model would be speed, which relates distances and time in a linear behaviour, which function is

s=\frac{d}{t}

So, if one ball has a speed of 3 meters per second, the function would be

3=\frac{d}{t}\\ d=3t

Where t represents the independent variable (horizontal axis) because it represents time, and d is distance, which is a dependent variable.

On the other hand, a non linear model could be population growth, which is an exponential beahaviour. It's not represented with a linear function, that's why is not linear.

An example of a function that represents population growth is

N(t)=N_{0} e^{rt}

Where t is time (horizontal axis), r is growth rate, N_{0} is the initial population and N(t) is the population after t time.

A specific example is

N(t)=12e^{0.012t}

At last, each function graph is attached, there you could see a lot better, the linear and non-linear behaviour.

UNO [17]3 years ago
3 0
While linear<span> equations are always straight, </span>nonlinear<span> equations often feature curves.</span>
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Answer:

Step-by-step explanation:

Given Parameters

Mean, x = 180

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Critical Value, t_\alpha, is given by

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Confidence Interval, \mu, is given by the formula

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }

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Critical Value, t_{\alpha/2}, is given by

t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093

Confidence Interval, \mu, is given by

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }

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Given data in the question:-

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