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3 years ago

13

Draw 2 cards from a standard deck of 52 cards. What is the probability that the first card is a queen and the second card is a k

ing?

1 answer:

bixtya [17]3 years ago

3 0

Answer:

4/663

Step-by-step explanation:

There are four queens. The probability that the first card is a queen is 4/52.

There are four kings. There are 51 cards left after the first card has been drawn, so the probability that the second card is a king is 4/51.

Therefore, the probability that the first card is a queen AND the second card is a king is:

P = 4/52 × 4/51

P = 4/663

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11. What method can be used to prove these two triangles are congruent?

zubka84 [21]

This would be SAS. If that’s wrong I’m sorry but it’s the only thing that makes sense, the dashes indicate they are the same and then the circle at the point also indicates that they’re the same so you have two sides and one angle that are the same, so it should be SAS.

5 0

2 years ago

There’s a picture .<br> Wich number below is a power of 6?

bekas [8.4K]

Hello from MrBillDoesMath!

Answer:

36

Discussion:

36 = 6^2 so is a power of 6

The other choices are multiples of 6 not powers of 6 (e.g. 24 = 6*4, 12 = 6*2, 18 = 6*3)

Tha nk you,

MrB

4 0

3 years ago

Y = x —6x+8 The equation above represents a parabola in the xy-plane. Which of the following equivalent forms of the equation di

forsale [732]

Answer:

$y=(x-4)(x-2)$

Step-by-step explanation:

Consider the given equation

$y=x^2-6x+8$

Factor form of a parabola: It displays the x-intercepts.

$y=a(x-p)(x-q)$ .... (1)

where, a is a constant and, p and q are x-intercepts.

So, we need to find the factored form of the given equation.

Splinting the middle term we get

$y=x^2-4x-2x+8$

$y=x(x-4)-2(x-4)$

$y=(x-4)(x-2)$ .... (2)

On comparing (1) and (2) we get

$p=4,q=2$

It means x-intercepts of the given parabola are 4 and 2.

Therefore, equivalent forms of the equation is y=(x-4)(x-2).

3 0

3 years ago

What two numbers add to 10 and multiply to -192? How do you find it?

Levart [38]

There is a way to do this algebraically using equations, but it is really messy so I am going to go over the way to do it. The way you do this is by finding whole numbers that multiply to make -192. Start with 1 and 192, 2 and 96, 3 and 64, etc and make the smaller number negative.
The reason why you make the smaller number negative is because we are looking for what numbers add together to make a positive number (10). If the bigger number was negative, then we would get a negative sum.
From what i could find, there aren't any whole numbers that satisfy this, and i doubt that the question would want you to find decimals for this.

The way to find this with equations (in short) is to have to variables, x and y, and make two equations. in this case x+y=10 and xy=-192. Solve one equation for a variable and then use substitution to plug that variable into the other equation. if you do this you get really long decimals.

7 0

3 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

$\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n$

$\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n$

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

3 years ago