Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.
Find the second decile of the following data set 24, 64, 25, 40, 45, 34, 14, 26, 28, 24, 58, 51 D2 =
Sergio039 [100]
Answer:
D2 = 24
Step-by-step explanation:
given is a data set as
24, 64, 25, 40, 45, 34, 14, 26, 28, 24, 58, 51
We have to find the second decile for the above
Arranging in ascending order we get
14, 24, 24, 25,26,28, 34, 40, 45, 51, 58, 64
2nd decile is equal to 20th percentile
20th percentile formula entry appearing which sorts 20% below and 80% above.
i.e. (20%*(14+1)) = 3rd entry
Thus 20th percentile is24.
D2 = 24
Answer:
The diagonal is increasing at the rate of 119/104cm/min of the given rectangle.
Step-by-step explanation:
Dimensions of the rectangle
Height = 5cm
Rate of base = 3/2 cm/min
Area = 60cm^2
We know the area of a rectangle of given by = base* Height
b*h = 60
b*5 = 60
b = 12cm
Applying Pythagoras theorem while drawing a diagonal to the rectangle
so our diagonal will be 13cm
Upon differentiating the area of the rectangle we get
b*h = A=60cm^2
using the chain rule of differentiation
h*db/dt + b*dh/dt = 0
b*dh/dt = -h*db/dt
12*dh/dt = -5*3/2
dh/dt = -5/8 cm//min
so the height of the rectangle is decreasing at the rate of -5/8cm/min
now we have all the measurements we need
b = 12 , db/dt = 3/2cm/min
h = 5 , dh/dt = -5/8 cm/min
Upon differentiating we get
2b*db/dt + 2h*dh/dt = 2D*dD/dt
b*db/dt + h*dh/dt = D*dD/dt
12*3/2 + 5*(-5/8) = 13*dD/dt
18 -25/8 = 13*dD/dt
= 13*dD/dt
dD/dt = 
Therefore the diagonal is increasing at the rate of 119/104cm/min of the given rectangle.
Probability that a student will play both is 7/30
Step-by-step explanation:
Total students = 30
No. of students who play basketball = 18
Probability that a student will play basketball = 18/30
= 3/5
No. of students who play baseball = 9
Probability that a student will play baseball = 9/30
= 3/10
No. of students who play neither sport = 10
Probability that a student will play neither sport = 10/30
= 1/3
To find :
Probability that a student will play both = p(student will play both)
No.of students who play sport = 30 - 10
= 20
Out of 20 students 18 play basketball and 9 play baseball.
So, some students play both the sports.
No. of students who play both sports = 18 + 9 - 20
= 7
p(student will play both) = 7/30
Probability that a student will play both is 7/30
Answer: 75%
Step-by-step explanation:
Height given :
69, 62, 73, 67, 63, 65, 73, 71, 70, 66, 59, 75
The total number of height given = 12
The height less than 73 are 69 , 62, 67 ,63 , 65 , 71 , 70 , 66 , 59
The total number of heights less than 73 = 9
Therefore : the percentage of these students who are shorter than 73 inches = total number of students shorter than 73/ total number of height x 100
= 9/12 x 100
= 75%
