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lora16 [44]
3 years ago
10

What is the equation of BD, simplified?

Mathematics
1 answer:
FromTheMoon [43]3 years ago
6 0

Third option is the correct answer.

Answer:

y = \bigg[ \frac{2b}{(2a - c)} \bigg]x - \bigg[ \frac{2bc}{(2a - c)} \bigg]

Step-by-step explanation:

y - y_1 = m(x - x_1) \\  \\ y - 0 =   \bigg[ \frac{2b}{(2a - c)} \bigg]  (x - c) \\  \\ y = \bigg[ \frac{2b}{(2a - c)} \bigg]x - \bigg[ \frac{2b}{(2a - c)} \bigg]c \\  \\  \purple { \boxed{ \bold{y = \bigg[ \frac{2b}{(2a - c)} \bigg]x - \bigg[ \frac{2bc}{(2a - c)} \bigg]}}} \\

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11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0
NISA [10]

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

Answer:

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

4 0
3 years ago
The quadrilateral shown is a rhombus. If AB = 17 and AE = 8, what is the measure of AC? A) 8 B) 12 C) 16 D) 24
Contact [7]

Answer:

C)16

Step-by-step explanation:

assuming the rhombus is named ABCD and E is where the diagonals meet.

Rhombus is a quadrilateral in which

  • All sides are equal
  • Opposite internal angles are equal
  • Diagonals bisect each other at right angles

As rhombus is  ABCD , AC is a diagonal which is bisected by E (According to the properties of Rhombus)

therefore AC will be twice of AE which is given to be 8

⇒ Answer is 16

5 0
3 years ago
Read 2 more answers
The weight W of an aluminum flatboat varies directly with the length L of the boat. If a 6​-foot boat weighs 102 ​pounds, then w
AnnZ [28]

Answer:

The 11 foot boat ways 187 pounds

Step-by-step explanation:

102 / 6 = 17

17 * 11 = 187

3 0
3 years ago
Tim's bank contains quarters, dimes and nickels. He has 3 more dimes than quarters and 6 fewer nickels than quarters. If he has
Artist 52 [7]
(x - 3) + (x - 6) + x = 63
x - 3 + x - 6 + x = 63
Combine like terms
3x - 9 = 63
Isolate the constant
3x - 9 + 9 = 63 + 9
3x = 72
Isolate the viable
3x / 3 = 72 / 3
x = 24
8 0
3 years ago
Write an expression for the area of the shaded region in its simplest form. Show all of your steps.
patriot [66]
The area of the shaded region will be the area of the rectangle minus the area of the white square inside of it:

((x+10)(2x+5)) - ((x+1)(x+1))

First, FOIL both of the areas separately:

(2x^2 + 5x + 20x + 50) - (x^2 + x + x + 1)

Simplify within the parentheses by adding like terms:

(2x^2 + 25x + 50) - (x^2 + 2x + 1)

Now, subtract one equation from the other:

2x^2 + 25x + 50
-x^2  -    2x  -   1
= x^2 + 23x + 49

This will be the equation for the area.
3 0
3 years ago
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