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Marizza181 [45]
4 years ago
8

20 points<3 please help

Mathematics
1 answer:
miv72 [106K]4 years ago
5 0

Hello there,

I hope so far you and your family are staying safe and healthy.

A) No, the events winning and playing are not independent of each other.

Why? Because as you can see, the probability of winning at home doesn't equal the probability of winning away.

B) No, the events losing and playing away are not independent either.

Why? Because of the same reason above. Their probabilities aren't equal.

Please let me know if you have questions about the answer. Best of luck with the rest of the semester!

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Find the minimum value of
viva [34]

Answer:

Find the minimum and maximum values of the objective function subject to the given constrants. Objective function: C= 2x + 3y

Constraints:

x>0

y>0

Comment: These two conditions tell you the answers are in the 1st Quadrant.

-----------------------------

x +y < 9

Graph the boundary line: y = -x+9

Solutions points are below the boundary line and in the 1st Quadrant.

8 0
3 years ago
Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
nordsb [41]

Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

4 0
3 years ago
What are the answers and why?
STALIN [3.7K]

Answer:

The selected answers are correct.

Step-by-step explanation:

The first step of the 3-step test for continuity is

  • check to see if the function is defined at the point. Here, the function h(-3) is defined as 5.

The second step of the 3-step test for continuity is

  • check to see if the limit exists at the point. Here, the limit is 2, coming at it either from the left or the right. (log6(36)=2; 16·2^-3=2)

The third step is

  • show the function value is the same as the limit at the point of interest. Here 5 ≠ 2, so there is a discontinuity at x=-3.
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79=27^x\\&#10;x=\log_{27}79
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