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
Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)
We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.
6x+3y=-12
Now just subtract it from the other equation.
6x+3y=-12
5x+3y=-6
***x=-6***
Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.
2(-6)+y=-4
-12+y=-4
y=8
Brainliest my answer if it helps you out?
24/64= 3/8
so the ratio is 3 to 8
Answer:
min = a_1
for i:= 2 to n:
if 
< min then min =
return min
Step-by-step explanation:
We call the algorithm "minimum" and a list of natural numbers 
So lets first set the minimum to 
min = a_1
now we want to check all the other numbers.
We can use a simple for loop, to find the minimum
min = a_1
for i:= 2 to n:
if < min then min =
return min
