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2 years ago

Write equation in standard form y-1= -1/4(x-6)

Mathematics

1 answer:

const2013 [10]2 years ago

6 0

The standard form:

$Ax+By=C$

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

$y-1=-\dfrac{1}{4}(x-6)$ use distributive property

$y-1=-\dfrac{1}{4}x+\dfrac{3}{2}$ multiply both sides by 8

$8y-8=-2x+4\cdot3$

$8y-8=-2x+12$ add 2x to both sides

$2x+8y-8=12$ add 8 to both sides

$\boxed{2x+8y=20}$

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Solving equations with variables on both sides 7+9x=6x-2

katrin2010 [14]

Answer:

Here

Step-by-step explanation:

We move all terms to the left:

7+9x-(6x-2)=0

We get rid of parentheses

9x-6x+2+7=0

We add all the numbers together, and all the variables

3x+9=0

We move all terms containing x to the left, all other terms to the right

3x=-9

x=-9/3

x=-3

3 0

2 years ago

Read 2 more answers

Plz help ill give you brainlist

Daniel [21]

Consider this option:
1. formula of perimeter is: P=2(a+b), where a & b - the sides of rectangle.
2. according to the condition 2(x+(x+4))<120; ⇒ x<28
answer: D. x<28

5 0

3 years ago

Find the transition matrix from B to B'. B = {(-1,2), (3, 4)), B' = {(1, 0), (0, 1)} STEP 1: Begin by forming the following matr

8_murik_8 [283]

Answer: The required matrix is

$T=\left[\begin{array}{ccc}-1&3\\2&4\end{array}\right] .$

Step-by-step explanation: We are given to find the transition matrix from the bases B to B' as given below :

B = {(-1,2), (3, 4)) and B' = {(1, 0), (0, 1)}.

Let us consider two real numbers a, b such that

$(-1,2)=a(1,0)+b(0,1)\\\\\Rightarrow (-1,2)=(a,b)\\\\\Rightarrow a=-1,b=2.$

Again, let us consider reals c and d such that

$(3,4)=c(1,0)+d(0,1)\\\\\Rightarrow (3,4)=(c,d)\\\\\Rightarrow c=3,d=4.$

Therefore, the transition matrix is given by

$T=\left[\begin{array}{ccc}-1&3\\2&4\end{array}\right] .$

Thus, the required matrix is

$T=\left[\begin{array}{ccc}-1&3\\2&4\end{array}\right] .$

7 0

2 years ago

I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0

2 years ago

2/x multiplied by what equals 1

Varvara68 [4.7K]

Answer:

Multiplied by2x

2/x ×2x =1

5 0

2 years ago