All categories

4 years ago

(Y = x - 3), (y = -3/5x + 5)

Mathematics

1 answer:

saul85 [17]4 years ago

7 0

Answer:

(5,2)

Step-by-step explanation:

i am unsure of what your question is but if you want the point of intersection of the two lines its (5,2)

You might be interested in

Fredrick manages a band and sings lead vocals the band. Charges $700 per concert. Fredrick receives $175 of that money,and he sh

kodGreya [7K]

Answer:

$111

Step-by-step explanation

Subtract 700-175 and that leaves $555 to split equally so then divide $555 by 5 and youll get $111

6 0

4 years ago

I have a 1.5% chance of winning when spinning a roulette. Approximately how many times would I have to spin the roulette to win?

fenix001 [56]

Answer:

66-67 times

Step-by-step explanation:

i would go with 67

you just divide 1.5 by a 100

3 0

3 years ago

A company designes a logo using a kite figure around the letter t. The logo is 13 centimeters wide and 16 centimeters tall. What

lara31 [8.8K]

13 X 16 =208cm^2
208 cm^2 is your answer

4 0

3 years ago

Santa is preparing for a

IrinaVladis [17]

Answer:

Cost of a package of celery = $20

Step-by-step explanation:

Let, x=carrots and y= celery.

So from the given data we get 2 equations

Equation 1]:-

$9x+12y=312$ ------------(for 9 package carrot and 12 package celery)

Equation 2]:-

$1x+3y=68$ ---------------(for 1 package carrot and 3 package celery)

After solving equation 2 we get,

$x=68-3y$ ------------------------(equation 3)

Now substituting equation 3 in equation 1,

$9\times(68-3y)+12y=312$

$612-27y+12y=312$

$612-15y=312$

$612-312=15y$

$300=15y$

∴ $y=\frac{300}{15}$

$y=20$ ------------------(equation 4)

Therefore cost of a package of celery = $20

6 0

4 years ago

Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

Answer:

Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

b) - With in migration process, 500 people are numbered. There will be after a long time,

$After\;inifinite\;period=[M]^n.[P]$

$Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$=\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

4 0

3 years ago