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lana66690 [7]
2 years ago
7

What is the area of the figure?

Mathematics
1 answer:
Finger [1]2 years ago
7 0

Answer:

Given a figure that is broken into a rectangle and triangle.

The dimensions of the rectangle are:

Base : 5 ft

Height : 1/3 ft

The dimensions of the triangle are :

Base : 3 upon 2/3 ft

Height : 2 ft

The area of the figure calculated below :-

Area of rectangle + Area of triangle

Area of rectangle + Area of triangleLB + 1/2BH

(5×1/3) + (1/3+3upon2/3×2)

The result is simplified below :-

5/3+11/3

5+11/3

16/3 ft

51/3 ft square.

The area of the figure is 5 and one third square feet.

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3 0
3 years ago
1) f(x) = 2x + 4, g(x) = 4x2 + 1; Find (g ∘ f)(0).
Sholpan [36]

Answer:

<h2>(g \: \circ \: f)(0) = 17</h2>

Step-by-step explanation:

f(x) = 2x + 4

g(x) = 4x² + 1

In order to find (g ∘ f)(0) we must first find

(g ° f )(x)

To find (g ° f )(x) substitute f(x) into g(x) that's for every x in g(x) replace it with f(x)

That's

<h3>(g \: \circ \: f)(x) = 4( ({2x + 4})^{2} ) + 1 \\  = 4(4 {x}^{2}  + 16x + 16) + 1 \\  =  {16x}^{2}  + 64x + 16 + 1</h3>

We have

<h3>(g \: \circ \: f)(x) =  {16x}^{2}  + 64x + 17 \\</h3>

Now to find (g ∘ f)(0) substitute the value of x that's 0 into (g ∘ f)(0)

We have

<h3>(g \: \circ \: f)(0) = 16( {0})^{2}  + 64(0) + 17 \\</h3>

We have the final answer as

<h3>(g \: \circ \: f)(0) = 17</h3>

Hope this helps you

8 0
3 years ago
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
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