Answer:
Step-by-step explanation:
slopes and equations:
find the equation thur ( 6,1 ) and (-2,-3)
find the slope m
m = (y2-y1) / (x2-x1 )
m = (-3 - 1) / (-2 -6)
m = -4 / -8
m = 1/2
now use the point-slope formula with our known slope
y-y1 = m(x-x1)
y-1 = 1/2(x-6)
y - 1 = 1/2x -3
y = 1/2x -3 +1
y = 
x -2
Find the equation parallel to y = 3x + 6 and thur (0,1)
Parallel means the same slope, the slope is 3 for the equation above.
use the slope-intercept formula again with the point given and the slope 3
y-1 = 3(x -0)
y - 1 = 3x
y = 3x +1
Find the equations perpendicular to 2x + y = 8 and the same y intercept as 4y = x + 3.
put both equations into proper form
y = -2x +6
y = 
x + 
perpendicular means reciprocal slope and change the sign, the 2nd equation has an intercept of , so
y = 1/2x +
there you go Amanda :)
Answer:
1. 80 yd
2. 10 and 10 of the other side
3. yes
Step-by-step explanation:
1. As we were asked to stay 240 feet away, and we will be on a soccer field and it has marked distances in yards we just have to pass our feet to yards.
1 ft = 1/3 yd
240 ft = 240/3 yd
240 ft = 80 yd
2. one would have to stand in yard 10 and the other in yard 10 but from the other side of the field
50-10 = 40
40 + 40 = 80 yd
3.
if they are diagonally the distance will be greater than 80, since the yards are parallel, because only the distance would remain only if they are on the same height
Answer:
n = 2
Step-by-step explanation:
1/2(n-4) - 3=3- (2n+3)
Expand the brackets.
1/2n-2-3=3-2n-3
Add or subtract the like terms.
1/2n-5=-2n
Add 2n and 5 on both sides.
1/2n+2n=5
Add like terms.
5/2n = 5
Multiply both sides by 2/5.
n = 5 × 2/5
n = 10/5
n = 2
Answer:
y = -1/3x - 5
Step-by-step explanation:
Perpendicular lines have negative reciprocal slopes, so the line is:
y = -1/3x + b
Finding the value of b, using the given point (-9, -2)
-2 = -1/3*(-9) + b
-2 = 3 + b
b = -2 - 3
b = -5
The line is
y = -1/3x - 5
Hopefully, this helps!
Answer:
P = 0,0012 or P = 0.12 %
Step-by-step explanation:
We know for normal distribution that:
μ ± σ in that range we find 68.3 % of all values
μ ± 2σ ⇒ 95.5 % and
μ ± 3σ ⇒ 99.7 %
Fom problem statement
We have to find (approximately) % of cars that reamain in service between 71 and 83 months
65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean
Probability between 65 and 71 is 49.75 %
On the other hand 74 is a value for mean plus 1, 5 σ and
74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).
Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87
Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75
P = 0,0012 or P = 0.12 %
