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Lubov Fominskaja [6]
3 years ago
5

Example 2: Combining Use of the Multiplication and Addition Rules

Mathematics
1 answer:
bulgar [2K]3 years ago
3 0

Answer:

The probability of getting 6's on both cubes is \frac{1}{36}.

The probability that the total score is at least 11 is \frac{1}{12}.

Step-by-step explanation:

Consider the provided information.

A red cube has faces labeled 1 through 6, and a blue cube has faces labeled in the same way.

Part (A) Both cubes show 6’s.

Probability of getting 6 on red cube is \frac{1}{6}

Probability of getting 6 on blue cube is \frac{1}{6}

Thus, the probability of getting 6's on both cubes is:

P(\text{Both 6's})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}

Hence, the probability of getting 6's on both cubes is \frac{1}{36}.

Part (B) The total score is at least 11.

The possible number of outcomes in which total score is at least 11 is:

Red shows 6 and Blue shows 5.

Blue shows 6 and Red shows 5.

Blue shows 6 and Red shows 6.

Thus, the probability of total score is at least 11.

P(\text{Total is at least 11})=\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}\\P(\text{Total is at least 11})=\frac{1}{12}

Hence, the probability that the total score is at least 11 is \frac{1}{12}.

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Answer:

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

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of y=2x^2-7x+6 and then evaluate it for x=4.

(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}

y'=4x-7+0               Simplifying and apply constant rule: (c)'=0

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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