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 #1
avatar+391 
+2
Apr 13, 2018
 #2
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+1
Apr 13, 2018
 #1
avatar+129839 
+2

This one gets ugly, ACG....but....it is "do-able"

 

If we position the bottom left vertex of the square at (0,0)  the circle has a center of (0,0)  and a radius of 4.....so.....the equation is  x^2 + y^2  = 16

Note....since AN  = 3...the intersection  of  NM  and the circle occurs at  x  = 3

So....  3^2 + y^2  = 16

9+ y^2  = 16

y  = √[16 - 9 ]   = √7

 

Now...if you can picture this.....let this intersection be the point,  Q

Draw  AQ  and  QD

 

Now....we want to find the area of the sector AQD

Note that the  sine of the central  angle  QAD  is NQ / AQ =   √7/4

So...the measure [in degrees]  of this angle can be expressed as  arcsin (√7/4)

So....the area of the sector is  (1/2)r^2 *arcsin (√7/4) (/360° }=

(1/2) 4^2* [arcsin (√7/4) / 360°]   =  8 [arcsin (√7/4) / 360° ]     (1)

Don't worry....this looks ugly, but it will eventually disappear !!!!

 

And the area of the triangle AQD  is  given by

(1/2) r^2 sin QAD   =

(1/2) 4^2 (√7/4)  =

2√7    (2)

 

So....the area of the sector - the area of the triangle  =  the area between the chord QD and the edge of the circle   =  (1) - (2)

 

And this is part of the white area NQD

And the other part of this white area is the triangle with base ND  and height  NQ  =

(1/2)ND * NQ   =  (1/2)(1)*√7  =  √7/2      (3)

 

So..... the "white"  area NQD  is given by  [ (1) - (2) ] + (3)  =

 8 [arcsin (√7/4) / 360° ]  - 2√7  +  √7/2  =

8 [arcsin (√7/4) / 360° ] -  3√7/2      (4)

 

So.....Area  "y"   =  Area of rectangle  MCDN  - (4)  =

MC * CD  - (4)  =

1 * 4  -  (4) =

4  -  [8 arcsin (√7/4) / 360° ] -  3√7/2 ] =

[  4 + 3√7/2   - 8 [arcsin (√7/4) / 360°  ]   =  (5)

 

Now  ...area  "x"   is   just the area  of the quater circle  - the white area,   (4)

The area of the quarater circle  is  just  (1/4)pi (4)^2)  = 4pi

So "x"   =

4pi - [8 [arcsin (√7/4) / 360° ] -  3√7/2]   =

4pi + 3√7/2 -  [8 arcsin (√7/4) / 360°]   (6)

 

So..... "x"  - "y"  =   (6) - (5)  =

 

  (4pi + 3√7/2 -  [8 arcsin (√7/4) / 360°] )   -  ( [  4 + 3√7/2   - 8 [arcsin (√7/4) / 360°  ] )  =

 

( 4pi  -  4)     units^2

 

See???...I told you it was ugly.... LOL!!!!

 

 

 

cool cool cool

Apr 13, 2018
 #2
avatar+5265 
+2

Part 1 involving the Spinner

 

Firstly, let's look at the first probability statement. We have an ample chance of landing the same spin again, so we won't be removing one blue for the next spin.

This fits the denominator 64, as that is 8*8. 

 

We first have a 5/8 chance to land on a blue, then we have a 3/8 chance to land on a red. 

\({5\over{8}}*{3\over{8}}\)is our equation (sorry for it being tiny, LaTeX is small)

 

In fraction multiplication, we multiply the top and bottom separately. 5*3 is 15, and as we said before 8*8 is 64

This leaves us with the fraction 15/64, which is true.

Finding this answer ALSO eliminates the third answer from being true, as it said that we remove a piece of the spinner if it is landed on, which is untrue.

Keep in mind that (red, blue) is the same as (blue, red)

-------------------------------------------------------------------------------

Now, let's look at the second option. 

 

The probability of (blue, blue) is \({5\over{8}}*{5\over{8}}\)

 

Now, let's multiply. 5*5 is 25, and 8*8 is 64. 

Therefore, this answer is true

-------------------------------------------------------------------------------

Finally, let's look at the fourth option. (Remember that the third option was not correct.)

 

The probability of (red, red) is \({3\over{8}}*{3\over{8}}\)

 

Now, let's multiply. 3*3 is 9, and 8*8 is 64.

9/64 does NOT reduce to 3/32. 

Therefore, this is a FALSE statement. 

 

The correct statements are Options 1 and 2.

 

-------------------------------------------------------------------------------

 

Part 2 involving Playing Cards

 

First, let's look at the part involving (Queen, Black Card)

 

First, this is a 4 in 52 or 1 in 13 chance of drawing a queen.

 

Then we draw a black card. 

The chance of this is 1 in 2.

 

We'll use the simplified numbers to form our equation.

\({1\over{13}}*{1\over{2}}\)

 

Now we multiply. 1*1=1, and 13*2=26

We now end with 1/26

-------------------------------------------------------------------------------

Now let's look at the probability of (Diamond, Ace)

 

First, the chance of drawing a Diamond in a set of 52 cards is 13 out of 52, or 1 in 4.

 

Then you have a 4 in 52 or 1 in 13 chance of drawing an Ace after the card is replaced.

 

Let's set up the equation: \({1\over{13}}*{1\over{4}} \)

 

Now, let's multiply. 1*1=1, and 13*4=52.

Therefore, our answer is 1/52.

-------------------------------------------------------------------------------

Thirdly, let's look at the probability of (Jack, 4)

 

First, the probability of drawing a Jack is 1 in 13.

 

Then, after replacement, the probability of drawing a 4 is 1 in 13.

 

We can make this easy on ourselves by squaring the fraction since they are the same. \({1\over{13}}^2\)

1^2=1, and 13^2=169

Therefore, our final answer is 1/169.

-------------------------------------------------------------------------------

Finally, let's look at the probability that we first draw a red card, and then a Club (Red card, Club)

 

The probability of drawing a red card is 1 in 2.

The probability of drawing a club is 1 in 4.

 

Now, let's set up the equation: \({1\over{2}}*{1\over{4}}\)

Now let's multiply. 1*1=1, and 2*4=8

 

Therefore, our answer is 1/8.

-------------------------------------------------------------------------------

 

Hope this helped and sorry for the long answer!

Apr 13, 2018
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Apr 13, 2018
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