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3 years ago

Helps me solve this problem please

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

X equals 0,4.........................

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On each round, Ann and Bob each simultaneously toss a fair coin. Let Xn be the number of heads tossed in the 2n flips which occu

CaHeK987 [17]

Answer:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

Step-by-step explanation:

In a coin toss the probability of tossing a head is 0.5 (50% head/50% tails)

If n is the number of rounds and 2n the number of coins tossed (one for each player), the probability of having m heads tossed is:

$R=\frac{2n!}{m!*(2n-m)!}$

R is the number of cases (combination of coins tossed) that gives a m number of heads. Each case has a probability of $P_{case}=0.5^{2n}$ so:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

For example, to toss 4 heads in 5 rounds:

n=5
2n=10
m=4

$P=\frac{10!}{4!*(10-4)!}*0.5^{10}$

$P=\frac{10*9*8*7*6!}{4!*6!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}=0.205$

8 0

3 years ago

Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exact C=.

Tpy6a [65]

Answer with Step-by-step explanation:

The given differential equation is

$(2x+5y)dx+(5x-4y)dy=0$

Now the above differential equation can be re-written as

$P(x,y)dx+Q(x,y)dy=0$

Checking for exactness we should have

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5$

$\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5$

As we see that the 2 values are equal thus we conclude that the given differential equation is exact

The solution of exact differential equation is given by

$u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)$

The value of $\phi (y)$ can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)

$\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c$