Question 1040348
1.Two dice are rolled. 
<pre>
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)  

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

(1,1)  (1,2)  (1,3)  <font color="red"><b>(1,4)</b></font>  (1,5)  (1,6)

(2,1)  (2,2)  <font color="red"><b>(2,3)</b></font>  (2,4)  (2,5)  (2,6)

(3,1)  <font color="red"><b>(3,2)</b></font>  (3,3)  (3,4)  (3,5)  (3,6)

<font color="red"><b>(4,1)</b></font>  (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 {{{4/36}}}.
That reduces to {{{1/9}}}
</pre>
b.at least 11? 
<pre>

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)  <font color="red"><b>(5,6)</b></font>

(6,1)  (6,2)  (6,3)  (6,4)  <font color="red"><b>(6,5)  (6,6)</font></b>

I count 3.  Therefore the probability is "3 times out of 36" or {{{3/36}}}.
That reduces to {{{1/12}}}
</pre>
c.less than 4?
<pre>
Let's color the ones red that have sum less than 4:

<font color="red"><b>(1,1)  (1,2)</b></font>  (1,3)  (1,4)  (1,5)  (1,6)

<font color="red"><b>(2,1)</font></b>  (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 {{{3/36}}}.
That reduces to {{{1/12}}}


  

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.

{{{drawing(150,200,-6,3,-9,3,
graph(150,200,-6,3,-9,3,x^2+4x-3), green(line(-10,-7,10,-7)) )}}}

We use the vertex formula:

The x-coordinate of the vertex of f(x) = ax²+bx+c is {{{-b/(2a)}}}

So the x-coordinate of the vertex of f(x) = x²+4x-3 is {{{-4/(2*1)}}} = -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&#8729;(-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</pre>