Help Question! Due Today

Find the x- and y- intercepts of the line. 2X+3y=-18

Anna71 [15]

To find the x-intercept, plug zero into y and solve for x.

2x+3(0)=-18

2x=-18
/2 /2

x=-9
To solve for the y-intercept, do the same thing, but instead plug zero into the x-intercept to solve for y.

2(0)+3y=-18
3y=-18
/3 /3

y=-6

So, your answer to your question is the y-intercept=-6, and the x-intercept=-9
Hope this helps!:)

6 0

3 years ago

Alan has to finish an exam within 120 minutes. He finished the exam in 80 minutes. What is the percentage of the time he took fo

drek231 [11]

Answer:

66.67%

Step-by-step explanation:

80/120 = 66.67

7 0

3 years ago

In how many ways can 100 identical chairs be distributed to five different classrooms if the two largest rooms together receive

Eva8 [605]

Answer:

There are 67626 ways of distributing the chairs.

Step-by-step explanation:

This is a combinatorial problem of balls and sticks. In order to represent a way of distributing n identical chairs to k classrooms we can align n balls and k-1 sticks. The first classroom will receive as many chairs as the amount of balls before the first stick. The second one will receive as many chairs as the amount of balls between the first and the second stick, the third classroom will receive the amount between the second and third stick and so on (if 2 sticks are one next to the other, then the respective classroom receives 0 chairs).

The total amount of ways to distribute n chairs to k classrooms as a result, is the total amount of ways to put k-1 sticks and n balls in a line. This can be represented by picking k-1 places for the sticks from n+k-1 places available; thus the cardinality will be the combinatorial number of n+k-1 with k-1, ${n+k-1 \choose k-1}$ .

For the 2 largest classrooms we distribute n = 50 chairs. Here k = 2, thus the total amount of ways to distribute them is ${50+2-1 \choose 2-1} = 51$ .

For the 3 remaining classrooms (k=3) we need to distribute the remaining 50 chairs, here we have ${50+3-1 \choose 3-1} = {52 \choose 2} = 1326$ ways of making the distribution.

As a result, the total amount of possibilities for the chairs to be distributed is 51*1326 = 67626.

7 0

3 years ago

The greatest comen factor of 21, 30,49

liq [111]

The greatest common factor for 21 is 7
For 30 is 15
49 is 7

I’m no totally sure but i think that is right

4 0

3 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