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1 year ago

According to the data, Vannessa mean quiz ...

Mathematics

1 answer:

damaskus [11]1 year ago

8 0

Answer: x = 108

Given the below equation

x + 13 1/2 = 121 1/2

Firstly, we need to convert the mixed fraction into an improper fraction

$\begin{gathered} x\text{ + 13}\frac{1}{2}\text{ = 121 }\frac{1}{2} \\ 13\frac{1}{2}\text{ = }\frac{(2\text{ x 13) + 1}}{2} \\ 13\text{ }\frac{1}{2}\text{ = }\frac{26\text{ + 1}}{2} \\ 13\frac{1}{2}\text{ = }\frac{27}{2} \\ 121\frac{1}{2}\text{ = }\frac{(2\text{ x 121) + 1}}{2} \\ 121\frac{1}{2}\text{ = }\frac{243}{2} \\ \text{Therefore, the new equation becomes} \\ x\text{ + }\frac{27}{2}\text{ = }\frac{243}{2} \\ \text{Isolate x} \\ x\text{ = }\frac{243}{2}\text{ - }\frac{27}{2} \\ \text{Common denominator = 2} \\ x=\text{ }\frac{243\text{ - 27}}{2} \\ x\text{ = }\frac{216}{2} \\ x\text{ = 108} \end{gathered}$

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Find by means of a vector( quantity) diagram . The resultant of two forces of 7 newton and 3 newton acting at right angle to one

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Answer:

The resultant of two forces of 7N and 3N acting at a right angle to one another is √58 N.

Step-by-step explanation:

Let the magnitude of the force of 7N be denoted as “F₁” on the x-axis and the magnitude of the force of 3N be denoted as “F₂” on the y-axis.

We are given that the two forces make an angle of 90° to one another as shown in the figure attached below.

Also, in the figure, we can see that the resultant force is denoted as “F”.

Now,

The magnitude of the resultant force can be calculated by using the Pythagoras theorem,

∴ F = √[F₁² + F₂²]

⇒ F = √[7² + 3²]

⇒ F = √[49 + 9]

⇒ F = √[58] N

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3 years ago

Read 2 more answers

Thaddeus has plotted several points on a scatter plot to investigate the relationship between two quantitative variables. He wan

Arturiano [62]

Answer:

no he does not

Step-by-step explanation:

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3 years ago

Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

$P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

$P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]$

Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

Transition matrix P raised to the power 5 (at k = 5)

$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]$

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0

4 years ago

On their way up the hill, Jack and Jill were stopped by their parents. It seems the two of them were trying to skip out on some

Katarina [22]

Answer:

77.14 minutes.

Step-by-step explanation:

Jack's time = 3 hours

Jill's time = 2.25 hours = 2 1/4 = 9/4 hours

Combined time = x

Rates :

Jack = 1/3 ; jill = 1 / 2.25 = 1 / 9/4 = 4 /9

Combined rate = 1/x

Setting up the equation :

1/3 + 4/9 = 1/x

Lcm of 3 and 9 = 9

(3 + 4) / 9 = 1/x

7/9 = 1/x

Cross multiply :

7x = 9

x = 9/7 hours

x = (9/7)* 60

x = 77.14 minutes

Combined time = 77.14 minutes.

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3 years ago

Lea and her father arrived at the scenic overlook 15 minutes before noon and left 12 minutes after noon using am or pm write the

Aneli [31]

Answer:

They arrived at am and left at pm

Step-by-step explanation:

What do you need to find?

We need to find the time when Lea and her father arrived at the scenic overlook and the time when they left.

What do you need to find first?

First we need to find the exact minutes they arrived and left. Then we have to determine whether it was am or pm

Steps:

1. Noting the time they arrived

They arrived 15 minutes before noon

2. Determining whether it was am or pm

Noon is 12 pm by convention

Midnight is 12 am by convention

3. Calculating the minutes for exact time

Each hour has 60 minutes:

60 - 15 = 45

4. Determining the time(am or pm) after subtracting 15 minutes

which means they arrived at 11:45 am (am is used for time before 12 at noon)

5. They left 12 minutes after the noon:

Noon time is 12 pm

6. Adding 12 minutes to 12pm

The exact time becomes 12:12pm

4 0

3 years ago