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laiz [17]
3 years ago
6

What is the effect on the graph of the function f(x) = 2^x when f(x) is replaced with f(x − 3)?

Mathematics
1 answer:
mel-nik [20]3 years ago
4 0

Graph of f(x-3) is compressed by a factor of  \frac{1}{8} horizontally of f(x).

<u>Step-by-step explanation:</u>

We have, the graph of f(x)= 2^{x} , on replacing f(x) by f(x-3) we get:

f(x-3)= 2^{x-3} = \frac{2^{x}}{2^{3}} = \frac{1}{8} 2^{x} = \frac{1}{8} f(x).Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of  \frac{1}{8} horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph : f(x-3) = \frac{1}{8} f(x).

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Suppose that each observation in a random sample of 100 fatal bicycle accidents in 2015 was classified according to the day of t
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Answer:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0:  proportion of fatal bicycle accidents in 2015 was the same for all days of the week

against the claim

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5) Calculations:

χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28

χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]

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6) Conclusion:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week

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