Answer:
The equation of the line is:
![y=2x-1](https://tex.z-dn.net/?f=y%3D2x-1)
Step-by-step explanation:
Given the points
Finding the slope between the points
![\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=%5Cmathrm%7BSlope%7D%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
![\left(x_1,\:y_1\right)=\left(-2,\:-5\right),\:\left(x_2,\:y_2\right)=\left(1,\:1\right)](https://tex.z-dn.net/?f=%5Cleft%28x_1%2C%5C%3Ay_1%5Cright%29%3D%5Cleft%28-2%2C%5C%3A-5%5Cright%29%2C%5C%3A%5Cleft%28x_2%2C%5C%3Ay_2%5Cright%29%3D%5Cleft%281%2C%5C%3A1%5Cright%29)
![m=\frac{1-\left(-5\right)}{1-\left(-2\right)}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B1-%5Cleft%28-5%5Cright%29%7D%7B1-%5Cleft%28-2%5Cright%29%7D)
![m=2](https://tex.z-dn.net/?f=m%3D2)
We know the slope-intercept form of the line equation
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
where m is the slope and b is the slope-intercept form
substituting the value m=2 and the point (-2, -5) to find the b-intercept
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
-5 = 2(-2) + b
b = -5+4
b = -1
Now, substituting m=2 and b=-1 in the slope-intercept form to get the equation of a line
y=mx+b
y=2x+(-1)
y=2x-1
Thus, the equation of the line is:
![y=2x-1](https://tex.z-dn.net/?f=y%3D2x-1)
to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those
![\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B2%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B6-2%7D%7B-1-0%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B-1%7D%5Cimplies%20-4%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-2%3D-4%28x-0%29%20%5C%5C%5C%5C%5C%5C%20y-2%3D-4x%5Cimplies%20%5Cblacktriangleright%20y%3D-4x%2B2%20%5Cblacktriangleleft)
now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)
![\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B2%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B6-2%7D%7B1-0%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B1%7D%5Cimplies%204%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-2%3D4%28x-0%29%20%5C%5C%5C%5C%5C%5C%20y-2%3D4x%5Cimplies%20%5Cblacktriangleright%20y%3D4x%2B2%20%5Cblacktriangleleft)
Problem 9
r+h = 9
SA = 2*pi*r^2 + 2*pi*r*h = 54pi
2*pi*r^2 + 2*pi*r*h = 54pi
2pi*r(r + h) = 54pi
r(r+h) = 27
r(9) = 27
9r = 27
r = 27/9
r = 3
r+h = 9
h = 9-r
h = 9-3
h = 6
Answers: r = 3 and h = 6 are the radius and height respectively.
==========================================================
Problem 10
d = diameter = 68 mm
r = radius = d/2 = 68/2 = 34 mm
SA = surface area of a sphere
SA = 4*pi*r^2
SA = 4*pi*34^2
SA = 4264pi
Answer: 4264pi square mm
==========================================================
Problem 11
Plug V = 3000 into the sphere volume formula and isolate r.
V = (4/3)pi*r^3
3000 = (4/3)pi*r^3
(4/3)pi*r^3 = 3000
4pi*r^3 = 3*3000
4pi*r^3 = 9000
r^3 = 9000/(4pi)
r = cube root( 9000/(4pi) )
r = ( 9000/(4pi) )^(1/3)
r = 8.947002 approximately
Now we can determine the surface area of this sphere.
SA = 4pi*r^2
SA = 4*pi*( 8.947002 )^2
SA = 1005.923451
Answer: 1005.923451 square feet approximately
==========================================================
Problem 12
We'll follow the same idea as problem 11, but in reverse.
SA = 4*pi*r^2
400pi = 4pi*r^2
r^2 = (400pi)/(4pi)
r^2 = 100
r = sqrt(100)
r = 10
Luckily we get a nice whole number for the radius r. Use it to find the volume.
V = (4/3)*pi*r^3
V = (4/3)*pi*10^3
V = (4000/3)pi
Answer: (4000/3)pi cubic inches exactly
==========================================================
Problem 13
The diagram is missing. I don't have enough info to be able to answer.
Answer:
32.33% probability of having at least 3 erros in an hour.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
The mean number of errors is 2 per hour.
This means that ![\mu = 2](https://tex.z-dn.net/?f=%5Cmu%20%3D%202)
(a) What is the probability of having at least 3 errors in an hour?
Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So
![P(X \leq 2) + P(X \geq 3) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%2B%20P%28X%20%5Cgeq%203%29%20%3D%201)
We want ![P(X \geq 3)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%203%29)
So
![P(X \geq 3) = 1 - P(X \leq 2)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%203%29%20%3D%201%20-%20P%28X%20%5Cleq%202%29)
In which
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.1353)
![P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.2707)
![P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2707)
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.1353%20%2B%200.2707%20%2B%200.2707%20%3D%200.6767)
![P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%203%29%20%3D%201%20-%20P%28X%20%5Cleq%202%29%20%3D%201%20-%200.6767%20%3D%200.3233)
32.33% probability of having at least 3 erros in an hour.