SOLUTION: 1.Two dice are rolled. What is the probability that the sum of the outcome is a.5? b.at least 11? c.less than 4? 2. Find the point on the graph of

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Question 1040348: 1.Two dice are rolled. What is the probability that the sum of the outcome is
a.5?
b.at least 11?

c.less than 4?
2. Find the point on the graph of f(x)=x^2+4x-3 where the tangent line is horizontal.

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
1.Two dice are rolled.
Here are all 36 possible outcomes when two dice are rolled:

(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)  


What is the probability that the sum of the outcome is
a.5?
Let's color the ones red that have sum 5:

(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)    

I count 4.  Therefore the probability is "4 times out of 36" or .
That reduces to 

b.at least 11?

Let's color the ones red that have sum of at least 11:

(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)

I count 3.  Therefore the probability is "3 times out of 36" or .
That reduces to 

c.less than 4?
Let's color the ones red that have sum less than 4:

(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)  

I count 3.  Therefore the probability is "3 times out of 36" or .
That reduces to 


  

2. Find the point on the graph of f(x)=x^2+4x-3 where the tangent line is horizontal.

The horizontal line through the vertex would be tangent at the
vertex, drawn in green below.



We use the vertex formula:

The x-coordinate of the vertex of f(x) = ax²+bx+c is 

So the x-coordinate of the vertex of f(x) = x²+4x-3 is  = -2.

The y-coordinate is found by substituting the x-coordinate of the
vertex in f(x) = x²+4x-3

f(2) = (-2)²+4∙(-2)-3 = 4-8-3 = -7

So the point on the graph of f(x)=x²+4x-3 where the tangent line is 
horizontal is the vertex (-2,-7). 


Edwin

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