You and a friend decide to sell cookies to raise

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

2 years ago

FIND THE VOLUME OF THE SPHERE!

Alexxx [7]

Answer:

v= 1.7in

Step-by-step explanation:

3 0

3 years ago

Find one arithetic means between 2 and 32

Murrr4er [49]

Answer:

17

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Your cousin is taking the train to Washington to visit her parents. She has at most $15.00 to spend on snacks and reading materi

erma4kov [3.2K]

I think 6 of each. I divided 15 by 2, got $7.50. You can buy exactly six magazines w that, and 6.5 granola bars. So six of each.

8 0

3 years ago

Does anyone know this?

Stels [109]

Every hexagon clearly has 6 sides. Nevertheless, every time you "glue" two hexagons together, you "lose" 2 sides to your count, because the sides where the two hexagons meet are not exterior sides anymore, and so they are not taken into account in our counting.

Also observe that with n hexagons you have n-1 points of contact between hexagons.

Since every hexagon has 6 sides and every gluing point takes away 2 sides, the number of exterior sides with n hexagons is

$s(n)=6n - 2(n-1) = 6n-2n+2=4n+2$

Let's plug some values for n:

$s(1) = 4\cdot 1 + 2 = 4+2 = 6$
$s(2) = 4\cdot 2 + 2 = 8+2 = 10$
$s(3) = 4\cdot 3 + 2 = 12+2 = 14$
$s(4) = 4\cdot 4 + 2 = 16+2 = 18$

You can check that these values are correct by counting the sides on the figure you have.

Finally, we can count the sides of a train with 10 hexagons by plugging n=10 in our formula:

$s(10) = 4\cdot 10 + 2 = 40 + 2 = 42$

Note: the numbers we've given are the number of sides that form the perimeter. So, the actual perimeters are the number of sides multiplied by the length of the side itself: if we let $s$ be the length of the side, the perimeters will be $6s,\ 10s,\ 14s,\ 18s$ for the first 4 trains, and $42s$ for the 10-hexagon train.

6 0

3 years ago