A national survey showed that 44% of college students drink. A professor provided his own survey by choosing a random sample of
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Answers:
- Exponential and increasing
- Exponential and decreasing
- Linear and decreasing
- Linear and increasing
- Exponential and increasing
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Explanation:
Problems 1, 2, and 5 are exponential functions of the form where b is the base of the exponent and 'a' is the starting term (when x=0).
If 0 < b < 1, then the exponential function decreases or decays. Perhaps a classic example would be to study how a certain element decays into something else. The exponential curve goes downhill when moving to the right.
If b > 1, then we have exponential growth or increase. Population models could be one example; though keep in mind that there is a carrying capacity at some point. The exponential curve goes uphill when moving to the right.
In problems 1 and 5, we have b = 2 and b = 1.1 respectively. We can see b > 1 leads to exponential growth. I recommend making either a graph or table of values to see what's going on.
Meanwhile, problem 2 has b = 0.8 to represent exponential decay of 20%. It loses 20% of its value each time x increases by 1.
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Problems 3 and 4 are linear functions of the form y = mx+b
m = slope
b = y intercept
This b value is not to be confused with the previously mentioned b value used with exponential functions. They're two different things. Unfortunately letters tend to get reused.
If m is positive, then the linear function is said to be increasing. The line goes uphill when moving to the right.
On the other hand if m is negative, then we go downhill while moving to the right. This line is decreasing.
Problem 3 has a negative slope, so it is decreasing. Problem 4 has a positive slope which is increasing.
Putting this as an arithmetic sequence gives:
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
![S_n=n/2 [2a+(n-1)d] = 784 000 \\n/2[2(150)+(n-1)200] = 784 000 \\n[300+200(n-1)=1 568 000 \\300n+200n^2-200n = 1 568 000 \\200n^2+100n- 1 568 000 = 0 \\2n^2 +n- 15680 = 0 \\n= 88.2...,-88.7](https://tex.z-dn.net/?f=S_n%3Dn%2F2%20%5B2a%2B%28n-1%29d%5D%20%3D%20784%20000%0A%5C%5Cn%2F2%5B2%28150%29%2B%28n-1%29200%5D%20%3D%20784%20000%0A%5C%5Cn%5B300%2B200%28n-1%29%3D1%20568%20000%0A%5C%5C300n%2B200n%5E2-200n%20%3D%201%20568%20000%0A%5C%5C200n%5E2%2B100n-%201%20568%20000%20%3D%200%0A%5C%5C2n%5E2%20%2Bn-%2015680%20%3D%200%0A%0A%5C%5Cn%3D%2088.2...%2C-88.7)
n has to be positive, so we get
n = <u>88.2 hours (3 s.f.)</u>
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
n has to be positive, so we get
n = <u>88.2 hours (3 s.f.)</u>