10m/s
32m/s
83m/s
None of these
Answer :
C
Solution :
At any instant x2+y2=52
Differentiating w.r.t. time 2xxdt+2ydydt=0
Here dxdt=2,dydt=u⇒u=83m/s
Answer:
BC = 4x-10
Step-by-step explanation:
is the midpoint of AC.
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
The density of the gold sample is 19.3g/cm³
Let the mass of the gold sample be represented by m
m = 579 g
Let the volume of the gold sample be represented by V
V = 30 cm³
Let the density of the gold sample be represented by ρ
The formula for the density is:
The density of the gold sample is 19.3g/cm³
Learn more here: brainly.com/question/17780219