each rear tire on an experimental airplane issupposed to be filled with a pressure of 40 pound per squareinch(psi).Let X denote the actual a

Question

each rear tire on an experimental airplane issupposed to be filled with a pressure of 40 pound per squareinch(psi).Let X denote the actual air pressure for the right tireand Y denote the actual air pressure for the left tire.Suppose that Xand Y are random varibles with the jointdensity
f(x,y)= k(x^2+y^2, 30<-x<50;
30<-y<50
0, elsewhere
a.) find k
b.) find P(30<-x<-40 and 40<-Y<50)
c.) Find the probability that both tires areunderfilled.
show all steps to solve…thx

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Cora 2 months 2021-10-08T22:31:59+00:00 1 Answer 0 views 0

Answers ( )

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    2021-10-08T22:33:46+00:00

    Answer:

    a. K = 3/3920000

    b. 0.26

    c. 0.189

    Step-by-step explanation:

    f(x,y) is a join distribution, so the following condition must be satisfied

    ∫∫ f(x,y) dydx {-∞ to ∞} {-∞ to ∞} = 1

    A. Finding K

    ∫∫ k(x² + y²) {30 to 50}{30 to 50} dydx = 1

    k ∫∫ (x² + y²) {30 to 50}{30 to 50} dydx= 1

    k ∫ (x²y + y³/3) {30 to 50}{30 to 50} dx= 1

    k ∫ (50x² + 50³/3 – 30x² – 30³/3) dx {30 to 50} = 1

    k ∫ (20x² + (50³ – 30³)/3) dx {30 to 50} = 1

    k ∫ (20x² + 98000/3) dx {30 to 50} = 1

    k(20x³/3 + 98000x/3) {30,50} = 1

    k(20 * 50³/3 + 98000 * 50/3 – 20 * 30³/3 – 98000 * 30/3) = @

    k(3920000)/3 = 1

    K = 3/3920000

    b.

    P(30<-x<-40 and 40<-Y<50) = ∫∫ k(x² + y²) {30 to 40}{40 to 50} dydx

    k ∫∫ (x² + y²) {30 to 40}{40 to 50} dydx

    k ∫ (x²y + y³/3) {30 to 40}{40 to 50} dx=

    k ∫ (50x² + 50³/3 – 40x² – 40³/3) dx {30 to 40}

    k ∫ (10x² + (50³ – 40³)/3) dx {30 to 40}

    k ∫ (10x² + 61000/3) dx {30 to 40}

    k(10x³/3 + 61000x/3) {30,40}

    k(10 * 40³/3 + 61000 * 40/3 – 10 * 30³/3 – 61000 * 30/3)

    k(980000/3)

    But k = 3/3920000

    So

    P(30<-x<-40 and 40<-Y<50) = 3/3920000 * 980000/3

    P(30<-x<-40 and 40<-Y<50) = 0.25

    c. Probability that both tries are undefined.

    This means that both pressure are between 30 and 40.

    So, we’ll solve for P(30<-x<-40 and 30<-Y<40)

    P(30<-x<-40 and 30<-Y<40) = ∫∫ k(x² + y²) {30 to 40}{30 to 40} dydx

    k ∫∫ (x² + y²) {30 to 40}{30 to 40} dydx

    k ∫ (x²y + y³/3) {30 to 40}{30 to 40} dx=

    k ∫ (40x² + 40³/3 – 30x² – 30³/3) dx {30 to 40}

    k ∫ (10x² + (40³ – 30³)/3) dx {30 to 40}

    k ∫ (10x² + 37000/3) dx {30 to 40}

    k(10x³/3 + 37000x/3) {30,40}

    k(10 * 40³/3 + 37000 * 40/3 – 10 * 30³/3 – 37000 * 30/3)

    k(740000/3)

    But k = 3/3920000

    So

    P(30<-x<-40 and 30<-Y<40) = 3/3920000 * 740000/3

    P(30<-x<-40 and 30<-Y<40) = 0.188775510204081

    P(30<-x<-40 and 30<-Y<40) = 0.189

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