Xp+yp=z. (ignore the period, any answers)
1 answer:
Xp + yp = z
You are solving for p :)
So, in order to find the value of p, you must get p to it's own side, and isolate it.
Since there are two "p's," you could change "xp + yp" to "p(x + y).
p(x + y) = z
Then, we must divide both sides by x + y
p(x + y) ÷ x + y = z ÷ x + y
Simplify.
p = z/x + y
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~Hope I helped!~
You are solving for p :)
So, in order to find the value of p, you must get p to it's own side, and isolate it.
Since there are two "p's," you could change "xp + yp" to "p(x + y).
p(x + y) = z
Then, we must divide both sides by x + y
p(x + y) ÷ x + y = z ÷ x + y
Simplify.
p = z/x + y
~Hope I helped!~
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The probability distribution for the face landing down = 4/15
<h3>What is the probability?</h3>The area of mathematics known as probability deals with numerical descriptions of how likely it is for an event to happen or for a claim to be true. A number between 0 and 1 is the probability of an event, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.
<h3>According to the given information:</h3>If we assign 1 probability x
then 2 has probability 1/2 x.
Similarly,
then 3 has probability = 1/2((1/2)*x) = (1/4)*x
so
then 4 has probability = 1/2((1/4)*x) = (1/8)*x
These probabilities need to sum to 1.
Hence:
or equivalently
(15/8)*x = 1
x = 8/15
the probability distribution for the face landing down = (1/2)*(8/15)
= 8/30
= 4/15
To know more about probability visit:
#SPJ4
I understand that the question you are looking for is:
A tetrahedral die has four faces, numbered 1-4. If the die is weighted in such a way that each number is twice as likely to land facing down as the next number (1 twice as likely as 2, 2 twice as likely as 3, and so on) what is the probability distribution for the face landing down?
Answer:
Step-by-step explanation:
<u>Area of ABCD = L * W</u>
<u>Area of EFGC:- L * W</u>
<u>Now, the area of the combind rectangles:-</u>
ABCD + EFGC