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3 years ago

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Tabitha has a deck of cards numbered 1−10. She picks one card, puts it back in the deck and then chooses a second card. What is

the probability that she picks an even number and then a 3?

2 answers:

Greeley [361]3 years ago

7 0

Tabitha has a deck of cards numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. There are 5 odd numbers 1, 3, 5, 7, 9 and 5 even numbers 2, 4, 6, 8, 10.

1. She picks one card from given 10 cards. The probability that she picks an even number is

$Pr_1=\dfrac{5}{10}=\dfrac{1}{2}.$

2. She puts first card back in the deck and then chooses a second card. The probability that she picks card with number 3 is

$Pr_2=\dfrac{1}{10}.$

3. Use the product rule to determine the probability that she picks an even number and then a 3:

$Pr=Pr_1\cdot Pr_2=\dfrac{1}{2}\cdot \dfrac{1}{10}=\dfrac{1}{20}=0.05.$

Answer: 0.05.

Valentin [98]3 years ago

3 0

First compute the whole number of the possibilities :
10*9=90.
Compute the number of favorable possibilities:
There is 5 even numbers. Once we pick an even number,
we pick 5. So the number of the favorable possibilities is 5.
Now we compute the probability like this:
$\frac{5}{90}$

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The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

2.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

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$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

3.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

4.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

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So the equation would be...

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