Help due in 30 min!!!!!!!!!!!!!!!

at a point P on the parabola x^2=4ay a normal PK is drawn. From vertex O, a perpendicular OM is drawn to meet the normal at M. S

saveliy_v [14]

<span>Answer: Its too long to write here, so I will just state what I did. I let P=(2ap,ap^2) and Q=(2aq,aq^2) But x-coordinates of P and Q differ by (2a) So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2) So Q=(2a(p-1), aq^2) which means, 2aq = 2a(p-1) therefore, q=p-1 then I subbed that value of q in aq^2 so Q=(2a(p-1), a(p-1)^2) and P=(2ap,ap^2) Using these two values, I found the midpoint which was: M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 ) then x = a(2p-1) rearranging to make p the subject p= (x+a)/2a</span>

6 0

3 years ago

Quadratic functions question! 7 pts!

Sedbober [7]

Answer:

Step-by-step explanation:

6,2

2,6

a is your answer

6 0

3 years ago

Which unit of length would be most appropriate to describe the length of your<br> foot?

ladessa [460]

Answer:

inches

Step-by-step explanation:

well depends on how big your foot is

8 0

3 years ago

You roll 2 dice. If the sum of the dice is: 2, 3, 4 you win $20. If the sum is 5, 6, 7, 8 you win $10. If the sum is 9, 10, 11 y

Elis [28]

Answer:

1) The expected value of the "winnings" is $(-0.97 $\overline 2$ )

2) The variance for the "winnings" is $0.57966

3) The standard deviation for the "winnings" is$0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

Step-by-step explanation:

1) The probability of having a sum of 2 = 1/6×1/6 = 1/36

The probability of having a sum of 3 = 1/6×1/6 = 1/36

The probability of having a sum of 4 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 5 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 6 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 7 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 8 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 9 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 10 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 11 = 1/6×1/6 = 1/36

The probability of having a sum of 12 = 1/6×1/6 = 1/36

The values are;

For 2, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 3, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 4, we have 1/18 × (20 - 5) = 0.8 $\overline 3$

For 5, we have 1/18 × (10 - 5) = 0.2 $\overline 7$

For 6, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 7, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 8, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 9, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 10, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 11, we have 1/36 × (-20 - 5) = -0.69 $\overline 4$

For 12, we have 1/36 × (-25 - 5) = -0.69 $\overline 4$

The expected value of the winnings is given as follows;

0.41 $\overline 6$ + 0.41 $\overline 6$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.41 $\overline 6$ + 0.41 $\overline 6$ + -1.3 $\overline 8$ -1.3 - 0.69 $\overline 4$ - 0.69 $\overline 4$ = -0.97 $\overline 2$

Therefore, the expected value = $-0.97 $\overline 2$ , which is one is expected to lose $0.97 $\overline 2$

2) Using Microsoft Excel, we have;

The variance for the "winnings", σ² = $0.57966

3) The standard deviation for the "winnings" = √σ² = √(0.57966) ≈ $0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

3 0

3 years ago

Describe the graph of the equation y = 6x.

Illusion [34]

y = 6x

y = (8.9) (9.8)

y = 6x (8.9)

y = 8.9x

4 0

3 years ago