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777dan777 [17]
2 years ago
11

Sketch the region bounded by the curves y=3x3, x+y=4, and y=0. Find the coordinates of the centroid.

Mathematics
1 answer:
Scilla [17]2 years ago
5 0
The graph of the region is attached.  The coordinates of the centroid are (5/3, 1).

The coordinates of the centroid are found by averaging the coordinates of the region;

Oₓ = (Aₓ+Bₓ+Cₓ)/3 = (0+1+4)/3 = 5/3
O(y) = (A(y) + B(y) + C(y)) = (0+3+0)/3=3/3=1

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BlackZzzverrR [31]

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

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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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Stefan's family rented a rototiller to prepare an area in their backyard for spring planting. The rental company charged an init
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Let h represent the hourly fee for renting the rototiller.

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43 + hx

If they paid $64 after renting the rototiller for 7 hours, the equation that models this problem would be

43 + 7h = 64

7h = 64 - 43 = 21

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8 0
2 years ago
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Answer:

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Answer:

1) ΔCBF ≅ ΔCDF by (SSS)

2) ΔBFA ≅ ΔDFE by (SAS)

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ΔCBF ≅ ΔCDF by Side Side Side (SSS) rule of congruency

2) Since DF ≅ BF and FA ≅ FE where ∠DFE = ∠BFA (alternate angles)

Therefore;

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EB and AD are the hypotenuse sides of ΔCBE and ΔCDA respectively

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EB = AD from  FE + BF = FA + DF

Where we also have BC ≅ DC

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BC and DC are the legs of ΔCBE and ΔCDA respectively

Then we have the following relation;

ΔCBE ≅ ΔCDA by Hypotenuse Leg (HL).

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