geno3141

Each one of the individual terms will have either a value of 1 or -1.

If a, b, and c are all positive, then so is abc, so the sum will be 1 + 1 + 1 + 1  =  4

If one of a, b, and c is negative, then so is abc, so the sum will be 0.

If two of a, b, and c are negative, then abc is positive, so the sum will be zero.

If all three are negative, then so is abc, so the sum will be -4.

The possible values are -4, 0, or 4.

For events to be independent one has no effect upon the other.

Thus, P(A) and P(A|B) must be the same; as also:  P(A) and P(B|A), P(B) and P(B|A) and P(B) and P(A|B).

Question #1:

1000 high school students -- this number goes into the far bottom right-hand corner -- the total of totals.

781 students own a cell phone -- this number goes into the bottom (total) slot of the first column.

-- you can now calculate the bottom (total) slot of the second column

133 you own a cell phone also own a PS4 -- this number goes into the top slot of the first column.

-- you can now calulate the second slot of the first column.

194 student don't own a cell phone or a PS4 -- this number goes into the the second slot of the second column.

-- you can now calculate all the missing numbers.

For problem #2, they are all correct except for part 5:

  What is the probability that a randomly student who prefers invisibility is an upperclassman?

This question assumes that the person prefers invisibility, so you are considering only those 43 students who

prefer invisibility.

Of these 43 students, 22 are upperclassmen, so the probability is 22/43.

8% of her salary was $4,000.00

   0.08 x salary  =  $4,000.00

              salary  =  $4,000.00 / 0.08

              salary  =  $50,000.00                               

Last year's salary was $3,000.00 less:

                  $50,000.00 - $3,000.00  =  $47,000.00

You can get a matched grey pair or a matched blue pair:

Grey:  6/14 x 5/13  =  15/91

Blue:  8/14 x 7/13  =  4/13

Adding these together to get the final answer:  43/91

Draw line segment AC, dividing the figure into two triangles.

In the top triangle, triangle(BPS) ~ triangle(BAC)   [side-angle-side for similarity]

Since BP = (2/3)^rds BA, the area of triangle(BPS)  =  (2/3 · 2/3)·triangle(BAC) =  4/9·triangle(BAC)

Similarly, the area of triangle(WDT) = 4/9·triangle(ADC).

Adding these two parts together, we get the fact that the unshaded area is 4/9^ths the quadrilateral.

However, we want the shaded area, so it will be 5/9^ths the quadrilateral.

(5/9)·180  =  100

Let's find a pattern:      a_n  =  a_{n - 1} - a_{n - 2}  

a₁  =  1                                        a₇  =  -1 - -2  =  1                         a₁₃  =  -1 - -2  =  1   

a₂  =  2                                        a₈  =  1 - -1  =  2                                 etc.

a₃  =  2 - 1  =  1                          a₉  =  2 - 1  =  1

a₄  =  1 - 2  =  1                          a₁₀  =  1 - 2  =  1

a₅  =  -1 -1 =  -2                          a11  =  -1 -1 =  -2

a₆  =  -2 - -1  =  -1                       a₁₂  =  -2 - -1  =  -1

Also notice that the sum of a₁ through a₆ is 0.

The first 16 groups  --  a₁ through a₉₆ have a sum of zero.

So, we just need to add the first 4 terms in the next group of 6 ...  

Can you finish it?

Getting the same answer as CPhill's but without trig.

By AA, all four triangles, the three smaller one and the full one are all similar.

Using CPhill's diagram, triangle(ABC) ~ triangle(FGC) ~ triangle(AFD).

Since triangle(ABC) is a right triangle, AB = sqrt(13).

Since the ratio of AC : CB : BA  =  2 : 3 : sqrt(13),

                            FC : CG : GF  =  2x : 3x : sqrt(13)·x          (for some value of x, that we can find later)

and                      DA : DF : FA  =  2y : 3y : sqrt(13)·y            (for some value of , that we can find later)

Since DFGE is a square:  FG = DF   --->   sqrt(13)·x  =  3y

                                                          --->   y  =  (sqrt(13) / 3) · x

This makes  AF  =  sqrt(13) · sqrt(13) / 3 · x   =   ( 13/3 )·x

AC  =  AF + FC   --->   2  =  (13/3)x + 2x   --->   6  =  13x + 6x   --->   6  =  19x   --->  x  =  6/19

Since FG  =  sqrt(13)·x   --->   FG  =  sqrt(13)·(6/9)  =  6·sqrt(13)/19

Same answer as CPhill's, so I feel confident!

First, place the problem in r·cis(theta) form.

  r  =  sqrt( x² + y² )  =  sqrt[ (5)² + (5·sqt(3))² ]  =  sqrt( 25 + 75 )  =  sqrt(100)  =  10

  theta  =  tan^-1( y / x )  =  tan^-1( 5·sqt(3) / 5 )  =  60^o

[ 10·cis( 60^o) ]⁷  =  10⁷ · cis( 7·60^o )  =  ...