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3 years ago

I need help with 4 and 5 what are the answers?

Mathematics

1 answer:

ira [324]3 years ago

7 0

Answer:

4. U = 54

5. C=80

Step-by-step explanation:

Basically you move the denominator to the other side and multiply. please take note of the signs(positve, negative)

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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

mamaluj [8]

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:

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

5 0

3 years ago

Without actual division prove that x4 +2x3 -2x2 +2x -3 is exactly divisible by<br> x2 +2x-3

Brut [27]

Answer:

Step-by-step explanation:

$\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}= \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1$