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3 years ago

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Mathematics

1 answer:

Sholpan [36]3 years ago

8 0

Answer:

Larry must mow approximately 25 lawns before he starts making a profit

Step-by-step explanation:

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A football player completes a pass 69.4% of the time. Find the probability that (a) the first pass he completes is the second

wlad13 [49]

Step-by-step explanation:

(a) If his second pass is the first that he completes, that means he doesn't complete his first pass.

P = P(not first) × P(second)

P = (1 − 0.694) (0.694)

P ≈ 0.212

(b) This time we're looking for the probability that he doesn't complete the first but does complete the second, or completes the first and not the second.

P = P(not first) × P(second) + P(first) × P(not second)

P = (1 − 0.694) (0.694) + (0.694) (1 − 0.694)

P ≈ 0.425

P = P(not first) × P(not second)

P = (1 − 0.694) (1 − 0.694)

P ≈ 0.094

6 0

3 years ago

Which three pieces of information contribute the most to your credit score?

Nadya [2.5K]

.....................................

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

How are comparisons between negative numbers different than comparisons between positive numbers?

lara [203]

On a number line, numbers always increase to the right and decrease to the left. Numbers to the right are greater than numbers to the left and numbers to the left are less than numbers to the right.

6 0

3 years ago

Where does the point (0, - 3) lie?

gregori [183]

An ordered pair is written like this, ( x, y ). In this case x = 0 and y = -3. On a graph the vertical line is the y-axis and the horizontal line is the x-axis. The origin is point ( 0, 0 ). To the left of the origin on the x-axis is the negative number line and to the right is the positive number line. On the y-axis, south of the origin is the negative number line and north is the positive number line. When you plot a point on a graph you do x first, so if x equals 1, you would move one right, -1, one left. IF y were to equal 2 then from the place where you are on the x-axis, 1, you would move two up, -2, two down. In this case x = 0 so you would stay at the origin, and y = -3 so you would move 3 down. So ( 0, -3 ) would lie negative y-axis. The answer is D.

3 0

3 years ago