Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3
Answer:
Brandon
Step-by-step explanation:
GIVEN: Branden and Pete each play running back. Branden carries the ball 
times for 
yards, and Pete has 
carries for 
yards.
TO FIND: Who runs farther per carry.
SOLUTION:
Total yards traveled by Brandon 
No. of times ball carried by Brandon 
Average yards per carry for Brandon 
Total yards traveled by Pete 
No. of times ball carried by Pete 
Average yards per carry for Pete
≅ 
As the number of yards per carry is higher for Brandon , therefore he runs farther per carry.
Answer:
X=-5y/13+90/13, Y=-13x/5+18
Step-by-step explanation:
You need to solve the equation for X and Y.
Solving for X:
13x+5y=90
Subtract 5y: 13x=-5y+90
Divide by 13: x=-5y/13+90/13
--
For Y:
13x+5y=90
Subtract 13x: 5y=-13x+90
Divide by 5: y=-13x/5+90/5
(Simplifies to -13x/5+18)
She had 18 to start with because when you multiply 4 by 2 it’s 8, so then you add 8 to 10 which is 18.
-8/7
6- -2= 8
3- -4=7
rise over run the slope of K is 7/8
perpendicular is the negative inverse of the slope so it ends up as -8/7
