Answer:
the probability is 2/9
Step-by-step explanation:
Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).
p(dime first) = 4/10 = 2/5
Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.
The joint probability of these two events occurring sequentially is the product of their probabilities:
p(dime then nickel) = (2/5)×(5/9) = 2/9
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<em>Alternate solution</em>
You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.
p(dime/nickel) = 20/90 = 2/9
Answer: Yes, the point (3,4) is a solution to the system.
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Proof of this:
Replace x with 3 and y with 4 in the first equation
x+y = 7
3+4 = 7
7 = 7
This confirms the first equation. Repeat for the second equation
x-2y = -5
3-2(4) = -5
3 - 8 = -5
-5 = -5
We get true equations for both when we plug in (x,y) = (3,4). This confirms it is a valid solution to the system of equations. It turns out it's the only solution to this system of equations. Visually, the two lines cross at the single location (3,4).
Answer:
16
Step-by-step explanation:
I dont really inderstand but i would say 16 because it is closer to adding up to 43 so yea
The answer is 74.5 if you saying 4 divided by 298 then that is the right answer
X(u, v) = (2(v - c) / (d - c) + 1)cos(pi * (u - a) / (2b - 2a))
y(u, v) = (2(v - c) / (d - c) + 1)sin(pi * (u - a) / (2b - 2a))
As
v ranges from c to d, 2(v - c) / (d - c) + 1 will range from 1 to 3,
which is the perfect range for the radius. As u ranges from a to b, pi *
(u - a) / (2b - 2a) will range from 0 to pi/2, which is the perfect
range for the angle. So, this maps the rectangle to R.
