You will need option A
AB = PQ
Answer:
326.6
Step-by-step explanation:
The formula for illuminance is given by
E = I / d^2
This formula only holds true for one-dimensional illuminance
The problem asks for the illuminance across the floor. We need to use two variables, x and y.
From Pythagorean Theorem
d^2 = x^2 + y^2
and from Trigonometry
x = d cos t
y = d sin t
The function for the illuminance can be represented by the composite function
E = I cos² t / x²
and
E = I sin² t / y²
The boundary of these functions is:
<span>0 < t < 8
So, the value of t must be in radians and not in degrees</span>
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:
m = (ps - b - uxs/t) / ux/t - p
Step-by-step explanation:
ux/t +b/m+s =p
multiplying throughout by (m+s) we get:
ux/t(m+s) + b = p(m+s)
open the brackets:
uxm/t + uxs/t + b = pm + ps
bring on one side of the equal sign all terms containg m, to make it the subject:
m(ux/t - p) = ps - b - uxs/t
m= (ps - b - uxs/t) / ux/t - p
