Answer:
V1 = 60 km/h
V2 = 40 Km/h
Step-by-step explanation:
The speed of an object is defined as
Speed = distance / time
Let
V1 be the speed of the faster car
V2 be the speed of the other car
t1 the time it took for the first car to arrive
t2 the time it took for the second car to arrive
d1 the distance traveled by first car
d2 the distance traveled by second car
We know thanks to the problem that
V1 = V2 + 20 Km/h
t1 = t2 - 1 hour
d1 = d2 = 120 Km
d1 = V1 * t1
d2 = V2* t2
V1 * t1 = V2* t2
V1* t1 = (V1 -20)*(t1 +1)
The system of equations
(V1 -20)*(t1 +1) = 120
V1 * t1 = 120
120 + (120/t1) -20*t1 = 140
(120/t1) -20*t1 = 20
Which gives,
t1 = 2
This means
V1 = 60 km/h
V2 = V1 - 20 Km/h = 40 Km/h
Answer:
Continuously
Step-by-step explanation:
Compounded continuously:
A = Pe^(rt)
A = 11,000 e^(0.0625 × 10)
A = 20,550.71
Compounded semiannually (twice per year):
A = P(1 + r)^t
A = 11,000 (1 + 0.063/2)^(2×10)
A = 11,000 (1 + 0.0315)^20
A = 20,453.96
Answer: 114
Step-by-step explanation:
To find the perimeter, you will find the distance around the shape by adding the numbers.
The is one missing side and to find that side you will subtract 10 from 38.
38 - 10 = 28
Now add the them.
28 + 5 + 10 + 26 + 10 + 35 = 114
Answer:
Example:
A bag contains 3 black balls and 5 white balls. Paul picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.
a) Construct a probability tree of the problem.
b) Calculate the probability that Paul picks:
i) two black balls
ii) a black ball in his second draw
Solution:
tree diagram
a) Check that the probabilities in the last column add up to 1.
b) i) To find the probability of getting two black balls, first locate the B branch and then follow the second B branch. Since these are independent events we can multiply the probability of each branch.
ii) There are two outcomes where the second ball can be black.
Either (B, B) or (W, B)
From the probability tree diagram, we get:
P(second ball black)
= P(B, B) or P(W, B)
= P(B, B) + P(W, B)
