Does anyone know the answer to this ?

Aleksandr [31]

Hey!

In order to simplify this equation, we'll first have to multiply both sides of the equation by v. This will give us v on its own.

Original Equation :
$\frac{-3}{v} = -6$

New Equation {Added Multiply Both Sides by V} :
$\frac{-3}{v} v=-6v$

Solution {New Equation Solved} :
$-3 = -6v$

Now we'll switch sides to get v on the left side of the equation which is generally where we always want the variables to be located in these types of equations.

Old Equation :
$-3 = -6v$

New Equation {Switched} :
$-6v=-3$

Now we'll divide both sides by v to get v on its own.

Old Equation :
$-6v = -3$

New Equation {Added Divide Both Sides by V} :
$\frac{-6v}{v} = \frac{-3}{v}$

Solution {New Equation Solved} :
$v = \frac{1}{2}$

So, this means that in the equation $\frac{-3}{v} =-6$ , $v = \frac{1}{2}$ .

Hope this helps!

- Lindsey Frazier ♥

3 0

3 years ago

Solve using the elimination method. show your work 2x - y = 6 x + y = 6

bekas [8.4K]

Answer:

x=4, y=2

Step-by-step explanation:

2x - y = 6

x + y = 6

Add these together to eliminate y

2x - y = 6

x + y = 6

-----------------

3x = 12

Divide by 3

3x/3 = 12/3

x=4

But we still need y

x+y =6

4+y =6

Subtract 4 from each side

4+y-4 = 6-4

y=2

7 0

3 years ago

The rainfall for one year was 35.8 inches. What was the approximate rainfall per month

Aloiza [94]

The rainfall for one month would be 2.983 inches per month.

7 0

3 years ago

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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

Is the square root of 13 a rational number or an irrational number?

abruzzese [7]

13 is not a perfect square so it is (irrational)

8 0

3 years ago

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