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babunello [35]
3 years ago
9

Find an exponential function with a horizontal asymptote y equals=2 whose graph contains the points ​(0,3​) and ​(1,6​).

Mathematics
1 answer:
Andrew [12]3 years ago
6 0
\bf y=a(b)^x\qquad \qquad (\stackrel{x}{0}~,~\stackrel{y}{3})\implies 3=a(b)^0\implies 3=a
\\\\\\
therefore\qquad y=3(b)^x\\\\
-------------------------------\\\\
(\stackrel{x}{1}~,~\stackrel{y}{6})\implies 6=3(b)^1\implies \cfrac{6}{3}=b^1\implies 2=b
\\\\\\
therefore\qquad y=3(2)^x

now, exponential functions have a horizontal asymptote at the x-axis, namely when y = 0, however, if you just move this one with a vertical translation of 2, then the horizontal asymptote will be at 2 instead.    y = 3(2)ˣ + 2
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In a recent year, the ACT scores for the math portion of the test were normally distributed, with a mean of 21.1 and a standard
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Answer:

a) P(X

And we can find this probability using the normal standard table or excel:

P(z

b) P(19

And we can find this probability with this difference:

P(-0.396

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-0.396

c) P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.925)=1- P(Z

d) We can consider unusual events values above or below 2 deviations from the mean

Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5

A value below 10.5 can be consider as unusual

Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7

A value abovr 31.7 can be consider as unusual

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X \sim N(21.1,5.3)  

Where \mu=21.1 and \sigma=5.3

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability using the normal standard table or excel:

P(z

Part b

P(19

And we can find this probability with this difference:

P(-0.396

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-0.396

Part c

P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.925)=1- P(Z

Part d

We can consider unusual events values above or below 2 deviations from the mean

Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5

A value below 10.5 can be consider as unusual

Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7

A value abovr 31.7 can be consider as unusual

6 0
3 years ago
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