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Andreyy89
3 years ago
6

In two or more complete sentences, describe the transformation(s) that take place on the parent function f(x) = 3^x to obtain th

e graph of f(x) = 3^-x + 1 +5 (5 not included in ^)
Mathematics
2 answers:
Korolek [52]3 years ago
6 0

first 3^x ----> 3^-x will reflect  3^x in the y axis  

then transformation to  3 ^(-x+1) will stretch  the graph parallel to y axis by a factor of 3.

Finally the - 5 will move the whole graph 5 units DOWNWARDS

not upwards. got it wrong because of that

kirill115 [55]3 years ago
5 0
First 3^x ----> 3^-x will reflect  3^x in the y axis 
then transformation to  3 ^(-x+1) will stretch  the graph parallel to y axis by a factor of 3.

Finally the + 5 will move the whole graph 5 units upwards.
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Answer:

a) P(X=1)=(5C1)(0.25)^1 (1-0.25)^{5-1}=0.39551

b)  P(X \leq 1) = P(X=0) +P(X=1)=0.23730+0.39551 =0.63281

c) P(X \geq 2) = 1-P(X

And for this case we can use the result from part b

P(X \geq 2) = 1-P(X

d) P(X \neq 1) since the random variable just takes values between 0 and 5 we can use the complement rule like this:

P(X \neq 1) = 1-P(X=1)= 1-0.39551=0.60449

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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The probability mass function for the Binomial distribution is given as:  

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Where (nCx) means combinatory and it's given by this formula:

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Part a

For this case our random variable X who represent the "number among the five who use the express checkout." follows X \sim Bin (n=5, p=0.25)

And we can find P(X=1) replacing on the mass function like this:

P(X=1)=(5C1)(0.25)^1 (1-0.25)^{5-1}=0.39551

Part b

For this case assuming that we want to find this probability P(X \leq 1) we can do this:

P(X \leq 1) = P(X=0) +P(X=1)

P(X=0)=(5C0)(0.25)^0 (1-0.25)^{5-0}=0.23730

P(X=1)=(5C1)(0.25)^1 (1-0.25)^{5-1}=0.39551

P(X \leq 1) = P(X=0) +P(X=1)=0.23730+0.39551 =0.63281

Part c

We can find P(2 \leq X) replacing on the mass function like this, using the complement rule:

P(X \geq 2) = 1-P(X

And for this case we can use the result from part b

P(X \geq 2) = 1-P(X

Part d

Assuming that we want to find this probability:

P(X \neq 1) since the random variable just takes values between 0 and 5 we can use the complement rule like this:

P(X \neq 1) = 1-P(X=1)= 1-0.39551=0.60449

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