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vazorg [7]
4 years ago
9

U = b - ak, solve for a

Mathematics
2 answers:
Lerok [7]4 years ago
5 0

Answer:  a = -\frac{u-b}{k}

kvv77 [185]4 years ago
4 0

Hey there!☺

Answer:\boxed{a=\frac{b-u}{k}}

Explanation:

u=b-ak

In our first step, we will flip the equation:

-ak+b=u

Now add -b to both sides:

-ak+b+-b=u+-b\\-ak=-b+u

In our last step, we will divide both sides by -k:

\frac{-ak}{-k}=\frac{-b+u}{-k}\\a=\frac{b-u}{k}

a=b-u/k is your answer.

Hope this helps!☺

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The scheduled arrival time for a daily flight from Boston to New York is 9:30 am. Historical data show that the arrival time fol
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Answer:

a) E(A)=\frac{0+50}{2}=25 min

Sd(A) =\sqrt{208.3333 min^2}=14.43 min

b) P(A>25) = 1-P(A\leq 25)=1-\frac{25-0}{50-0}= 1-0.5 = 0.5

Step-by-step explanation:

Part a

Let A the random variable that represent "The arrival time (minutes) for a daily flight from Boston to New York ". And we know that the distribution of A is given by:

A\sim Uniform(a=0 ,b=50)

We select our starting point a=0 representing 9:05 AM and the amount of minutes between 9:05 am to 9:55 am are 50 ans for this reason b =50

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{0+50}{2}=25 min

And the variance is given by:

Var(A)= \frac{(b-a)^2}{12}=\frac{(50-0)^2}{12}=208.3333 min^2

And the standard deviation by definition is just the square root of the variance:

Sd(A) =\sqrt{208.3333 min^2}=14.43 min

Part b

If we convert 9:30 AM to our scale we know that we have 25 minutes between 9:05 AM and 9:30 AM.

For this case we want to find this probability: P(A>25)

And we can find this probability using the complement rule like this:

P(A>25) = 1-P(A\leq 25)

And we can use the cumulative distribution function given by:

F(A) = \frac{A-a}{b-a}

And we got:

P(A>25) = 1-P(A\leq 25)=1-\frac{25-0}{50-0}= 1-0.5 = 0.5

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