Step-by-step explanation:
1. P(light OR domestic) = P(light) + P(domestic) − P(light AND domestic)
P(light OR domestic) = 0.62 + 0.70 − 0.55
P(light OR domestic) = 0.77
2. P(light AND not domestic) = P(light) − P(light AND domestic)
P(light AND not domestic) = 0.62 − 0.55
P(light AND not domestic) = 0.07
3. P(light GIVEN not domestic) = P(light AND not domestic) / P(not domestic)
P(light GIVEN not domestic) = 0.07 / (1 − 0.70)
P(light GIVEN not domestic) = 0.233
4. Two events are independent if P(A) × P(B) = P(A and B).
P(light) × P(domestic) = 0.62 × 0.70 = 0.434
P(light AND domestic) = 0.55
Therefore, the type and location are not independent.
Answer:
3.13 kilograms used for the second cake
Step-by-step explanation:
because the fractions have different denominators you have to multiply the denominators together to figure out what to change them too. but instead you can just figure out what the decimal for each is. 1/3 is 0.33, and 4/5 is 0.80. all you do is add these together to get 1.13. so now all you do is add 2+1.13 to get 3.13.
6.375 is greater than 6.333
Answer:
OH MY GOD I NEEED HELP WITH QUESTION TOOOOOOO
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.
