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marta [7]
2 years ago
9

Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile

to get ready to start. Assume that the joint density for(X,Y) is given by fxy (x,y) = c(4x + 2y +1) ; 0 < x< 40 and 0 < y <2.
(a) Find the value of c that makes this a density.
(b) Find the probability that on a randomly selected day theair temperature will exceed 20 degrees C and it will take at least1 minute for the car to be ready to start.
(c) Find the marginal densities for X and Y.
(d) Find the probability that on a randomly selected day itwill take at least one minute for the car to be ready tostart.
(e) Find the probability that on a randomly selected day theair temperature will exceed 20 degrees C.
(f) Are X and Y independent? Explain on mathematicallybasis.
Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
4 0

Answer:

Step-by-step explanation:

Given f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y

a)

we know that \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1

therefore \int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1

on integrating we get

c=(1/6640)

b)

P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

on doing the integration we get

                        =0.37349

c)

marginal density of X is

f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx

on doing integration we get

f(y)=\frac{(y+40.5)}{83}

d)

P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

solve the above integration we get the answer

e)

P(X>20, 0

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

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Answer:

The positive graph is on the points  (-5,-4) and (2,5)

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Step-by-step explanation:

As it seems that fgraph shows positive in the case when it is over x-axis, and negative when it is below y-axis

As on the interval (-1,5)

The graph is below x-axis i.e. on (-1,2)

And, over x-axis i.e. on (2,5)

At this point Lonzell’s  is not correct

As on the interval (-5,-1)

The graph is below x-axis i.e. on (-4,-1)

And, over x-axis i.e. on (-5,4)

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The positive graph is on the points  (-5,-4) and (2,5)

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Answer:

  (3/4)a

Step-by-step explanation:

The angle at K is 120°, so the angle at L is its supplement: 60°. That makes triangle FKL an equilateral triangle with a base of FL = a. The vertex at K is centered over the base, so is a/2 from G.

The midsegement length is the average of GK and FL, so is ...

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Triangles R S T and V U T are connected at point T. Angles R S T and V U T are right angles. The length of side R S is 12 and th
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Answer:

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Step-by-step explanation:

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See the attachment for better explanation.

6 0
2 years ago
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Statements A, B, and C are true.

Step-by-step explanation:

Step 1:

A cone, a cylinder and a sphere all have radii of 3 inches. The cone and cylinder have heights of 2 inches.

The volume of the cone, V=\pi r^{2} \frac{h}{3}. Here r = 3 inches and h = 2 inches.

So the volume of the cone  =\pi r^{2} \frac{h}{3} = \pi (3^{2}) \frac{2}{3} = 18.85 cubic inches.

The volume of the cylinder, V=\pi r^{2} h. Here r = 3 inches and h = 2 inches.

So the volume of the cylinder = =\pi r^{2} h= \pi  (3^{2}) (2) = 56.55 cubic inches.

The volume of the sphere, V=\frac{4}{3} \pi r^{3}. Here r = 3 inches.

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Step 2:

Now we check to see which statements are true.

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B. About 18.8 cubic inches of catnip will fit inside a cone-shaped toy so it true.

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D. No shape holds 169.6 cubic inches, so D is false.

E. The toys shaped like a cone and a cylinder do not hold the same amount of catnip in them. So it is false.

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in the drawing six out of every 10 tickets are winning tickets of the winning tickets one out of every three awards is a larger
Dafna11 [192]

Answer: P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

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Number of winning tickets =6

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P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

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