Answer:
x = -7 2/3, y = 1 1/3 and z = 5 1/3.
Step-by-step explanation:
2x+4y+3z=6 ..... 1
x-2y+z=-5 ...... 2
-x-3y-2z=-7 .......3
Add equations 2 and 3 to eliminate x:
-5y - z = -12 .....4
Multiply equation 2 by - 2:
-2x + 4y - 2z = 10
Add this to equation 1:
8y + z = 16 ........ 5
Now add equation 4 to equation 5:
3y = 4
y = 4/3 = 1 1/3.
Now find z by substituting for y in equation 4:
-5(4/3) - z = -12
z = 12 - 20/3
z = 36/3 - 20/3 = 16/3 = 5 1/3.
Finally, we find x by substituting for y and z in equation 1:
2x + 4*4/3 + 3*16/3 = 6
2x = 6 - 16/3 - 16
2x = 18/3 - 16/3 - 48/3 = -46/3
x = 23/3 = 7 2/3.
We are looking to find P(X>60 students)
X is normally distributed with mean 50 and standard deviation 5
We need to find the z-score of 60 students
To find the probability of P(Z>2), we can do 1 - P(Z<2)
So we read the probability when Z<2 which is 0.9772, then subtract from one we get 0.0228
The number of students that has score more than 60 is 0.0228 x 1000 = 228 students
<span>Sphere: (x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
Intersection in xy-plane: (x - 4)^2 + (y + 12)^2 = 36
Intersection in xz-plane: DNE
Intersection in yz-plane: (y + 12)^2 + (z - 8)^2 = 84
The desired equation is quite simple. Let's first create an equation for the sphere centered at the origin:
x^2 + y^2 + z^2 = 10^2
Now let's translate that sphere to the desired center (4, -12, 8). To do that, just subtract the center coordinate from the x, y, and z variables. So
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 10^2
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 100
Might as well deal with that double negative for y, so
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
And we have the desired equation.
Now for dealing with the coordinate planes. Basically, for each coordinate plane, simply set the coordinate value to 0 for the axis that's not in the desired plane. So for the xy-plane, set the z value to 0 and simplify. So let's do that for each plane:
xy-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (0 - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (-8)^2 = 100
(x - 4)^2 + (y + 12)^2 + 64 = 100
(x - 4)^2 + (y + 12)^2 = 36
xz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (0 + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + 12^2 + (z - 8)^2 = 100
(x - 4)^2 + 144 + (z - 8)^2 = 100
(x - 4)^2 + (z - 8)^2 = -44
And since there's no possible way to ever get a sum of 2 squares to be equal to a negative number, the answer to this intersection is DNE. This shouldn't be a surprise since the center point is 12 units from this plane and the sphere has a radius of only 10 units.
yz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(0 - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(-4)^2 + (y + 12)^2 + (z - 8)^2 = 100
16 + (y + 12)^2 + (z - 8)^2 = 100
(y + 12)^2 + (z - 8)^2 = 84</span>
they are giving you the answer AB=AC, so what that means is that the angles are congruent to each other so if angle AC=15 and angle DC=5 all the angles will be the same.
It’s either subtract 75 from both sides or divide both sides by 1/2 but I’m pretty sure it’s subtract 75 from both sides
