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3 years ago

If you roll a standard 6 sided die, whats the probability that you get a 1,2,3,5, or 6?

Mathematics

2 answers:

Papessa [141]3 years ago

6 0

Answer:

5/6

Step-by-step explanation:

1 has a 1/6 chance
2 has a 1/6 chance
3 has a 1/6 chance
5 has a 1/6 chance
6 has a 1/6 chance
1, 2, 3, 5, or 6 has a 1/6+1/6+1/6+1/6+1/6=5/6 chance

inna [77]3 years ago

3 0

Answer:

18.3% chance it will land on that particular side

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21. Randy owns a computer store. In 1990, he sold 150 monitors. In 2000, he sold 900

lutik1710 [3]

Answer:

$y=75x+150$

Step-by-step explanation:

Let $y$ represent the number of monitors sold.

Given:

$x$ is the number of years since 1990.

So, for the year 1990, $x=0$ .

For the year 2000, 10 years passed. So, $x=10$ .

Now, monitors sold in 1990 are 150. So, at $x=0,y=150$

Monitors sold in 2000 are 900. So, at $x=10,y=900$

Thus, the two points are $(0,150)$ and $(10,900)$ .

The slope of a line with points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given as:

Slope, $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the points $(0,150)$ and $(10,900)$ , the slope is given as:

Slope, $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ m=\frac{900-150}{10-0}\\ m=75$

Now, the y-intercept is the point where $x=0$ . So, the point $(0,150)$ has $x$ value as 0.

So, the y-intercept is 150.

Equation of a line in slope-intercept form is given as:

$y=mx+b$

Where, $m$ is the slope and $b$ is the y intercept.

Here, $m=75$ and $b=150$ .

Therefore, the equation that represents the above data is given as:

$y=75x+150$

7 0

3 years ago

The designer also programs a bird with a path that can be modeled by a quadratic function. The bird starts at the vertex of the

Nady [450]

Answer:

-1.2

Step-by-step explanation:

Given that the designer also programs a bird with a path that can be modeled by a quadratic function.

The bird starts at the vertex of the path at (0, 20) and passes through the point (10, 8).

If we treat this curve as line joining these two points then we can find the slope by the formula

Slope = change in y coordinate/change in x coordinate

Here the points given are

(0,20) and (10,8)

$Change in y coordinate = 8-20 = -12\\Change in x coordinate = 10-0 = 10\\Slope = -1.2$

Slope of the line that represents the turtle's path

=-1.2

6 0

3 years ago

12x - 2y = -1<br>+ 4x + 6y= -4<br>

svp [43]

Answer:

x = -7/40 , y = -11/20

Step-by-step explanation:

Solve the following system:

{12 x - 2 y = -1 | (equation 1)

4 x + 6 y = -4 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{12 x - 2 y = -1 | (equation 1)

0 x+(20 y)/3 = (-11)/3 | (equation 2)

Multiply equation 2 by 3:

{12 x - 2 y = -1 | (equation 1)

0 x+20 y = -11 | (equation 2)

Divide equation 2 by 20:

{12 x - 2 y = -1 | (equation 1)

0 x+y = (-11)/20 | (equation 2)

Add 2 × (equation 2) to equation 1:

{12 x+0 y = (-21)/10 | (equation 1)

0 x+y = -11/20 | (equation 2)

Divide equation 1 by 12:

{x+0 y = (-7)/40 | (equation 1)

0 x+y = -11/20 | (equation 2)

Collect results:

Answer: {x = -7/40 , y = -11/20

6 0

3 years ago

Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint. (Hint: Let (x,y) be the unkno

melomori [17]

Mid point formula =
(x1/2+x2/2, y1/2+y2/2)

x=
x+0 = 8
/2
x+0 = 8 x 2
x = 16

y=
y+(-2) = 1
/2
y-2 =1 x2
y=2+2
y=4

(16,4)

(In this case, the midpoint (x ,y) will be represent as the answer)

7 0

3 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

3 years ago