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lara31 [8.8K]
3 years ago
9

Which input value produces the same output value for the two functions on the graph?

Mathematics
2 answers:
xeze [42]3 years ago
8 0
X = 4 will give you the same output no matter what graph you are using. 
Elena-2011 [213]3 years ago
3 0

Answer:

x=4

Step-by-step explanation:

WE need to find out which input value x value gives the same output value y for the two function f(X) and g(X) on the graph

We look at the point of intersection of both f(x) and g(x)

both graphs intersects at a point (4,3)

It means for the input value x=4 both f(x) and g(x) has same output 3

So input value x=4 produces the same output value for the two functions on the graph.

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A group of students were surveyed to find out if they like building snowmen or skiing as a winter activity. The results of the s
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Answer:

The answer could be easier if you make a ratio and actual graph

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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

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2 years ago
-7a⁵· 6a write the product in simplest form ?
MariettaO [177]

Answer:

= -42a⁶

Step-by-step explanation:

-7a⁵. 6a = -7a⁵ × 6a

Therefore;

We multiply; -7 × 6 = -42

then; a⁵ × a = a⁶

Thus; we have -42 a⁶

Hence; <u>-7a⁵ × 6a = -42 a⁶</u>

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3 years ago
Read 2 more answers
L1 : y = 2x , (2) find the equation of the line L2 perpendicular to L1 passing through the point P = (1, 2).
just olya [345]

Answer:

<h2>2y+x = 5</h2>

Step-by-step explanation:

Given the line L1 as y = 2x perpendicular to an unknown line L2 passing through the point P = (1, 2), we are to find the equation of line L2. to find the equation of the line L2, we will use the point-slope equation of a line expressed as y-y₀ = m(x-x₀)

m is the slope of the unknown line

(x₀, y₀) is the given point.

First is to get the slope of the known line:

comparing the line L1: y = 2x with the standard equation of the line y = mx+c, it can be seen that m = 2

Then we will calculate the slope of the required line.

Since L1 is perpendicular to L2, the product of their slope will be -1 i.e

mm₁ = -1 where m₁ is the slope of the required line L2.

Given m =2

m₁ = -1/m

m₁ = -1/2

Finally we will calculate the equation of line L2 by substituting the slope of line L2 and the point in the point slope equation above;

y-y₀ = m(x-x₀)

Given (x₀, y₀) = (1,2) and m₁ = -1/2

y-2 = -1/2(x-1)

open the parenthesis

y-2 = -x/2+1/2

multiply through by 2:

2y-4 = -x+1

2y+x = 1+4

2y+x = 5

<em>Hence the equation of the line L2 is 2y+x = 5</em>

<em></em>

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2 years ago
Each side of a square classroom is 8 meters long. The school wants to replace the carpet in the classroom with new carpet that c
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