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3 years ago

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A standard deck of has 52 cards, and contains 13 cards of each of the four suits: diamonds, spades, hearts, and clubs. What is t

he probability that you choose a spade, given that you have already chosen a diamond (and have not replaced it)?

1 answer:

Verizon [17]3 years ago

7 0

Answer: $\frac{13}{51}$

Step-by-step explanation:

Given: Number of cards in a standard deck = 52

If we already chosen a diamond card, then the number of cards left in the deck =52-1=51

The number of spades remains same = 13

Now, the probability that you choose a spade, given that you have already chosen a diamond is given by :-

$P(Spade|diamond)=\frac{13}{51}$

Hence, the the probability that you choose a spade, given that you have already chosen a diamond $=\frac{13}{51}$

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The tangent line to the given curve at the given point is $y=9x-26$ .

Step-by-step explanation:

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$(y=2x^2-7x+6)'$ Differentiate the equation.

$(y)'=(2x^2-7x+6)'$ Differentiate both sides.

$y'=(2x^2)'-(7x)'+(6)'$ Sum/Difference rule applied: $(f(x)\pmg(x))'=f'(x)\pm g'(x)$

$y'=2(x^2)'-7(x)'+(6)'$ Constant multiple rule applied: $(cf)'=c(f)'$

$y'2(2x)-7(1)+(6)'$ Applied power rule: $(x^n)'=nx^{n-1}$

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$y'=4x-7$ Simplify.

Evaluate y' for x=4:

$y'=4(4)-7$

$y'=16-7$

$y'=9$ is the slope of the tangent line.

Point slope form of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Insert 9 for $m$ and (4,10) for $(x_1,y_1)$ :

$y-10=9(x-4)$

The intended form is $y=mx+b$ which means we are going need to distribute and solve for $y$ .

Distribute:

$y-10=9x-36$

Add 10 on both sides:

$y=9x-26$

The tangent line to the given curve at the given point is $y=9x-26$ .

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function $f(x)=2x^2-7x+6$ at $x=4$ (the slope of the tangent line at $x=4$ ):

$\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}$

$\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}$ We are given f(4)=10.

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

$2x^2-7x-4=(x-4)(2x+1)$

Let's check this with FOIL:

First: $x(2x)=2x^2$

Outer: $x(1)=x$

Inner: $(-4)(2x)=-8x$

Last: $-4(1)=-4$

---------------------------------Add!

$2x^2-7x-4$

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

$\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}$

$\lim_{x \rightarrow 4}(2x+1)$

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

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