There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
2*2*2*3*5*5
or
2^3 *3* 5^2
Step-by-step explanation:
600 = 60*10
These are not prime so keep going, breaking 60 and 10 down
= 6*10 * 5*2
6 and 10 are not prime so keep going ( 5 and 2 are prime)
= 3*2 * 5*2 * 5*2
All of the numbers are prime, so we normally write them in order from smallest to largest
2*2*2*3*5*5
We can write with exponents
2^3 *3* 5^2
Answer:
(4,1)
Step-by-step explanation:
x+y=5
x- y=3
Add the equations together to eliminate y
x+y=5
x- y=3
--------------------
2x = 8
Divide each side by 2
2x/2 = 8/2
x = 4
Now we can find y
x+y = 5
4+y = 5
Subtract 4 from each side
4+y-4 = 5-4
y =1
Simplification/ 15x^6-20x^5+10x^4
Answer:
1yr 47,000, 2nd yr 48,880, 3rd yr 51,617.28, 4th yr 57,346.7981 or 57,346.80
Step-by-step explanation:
1yr 47,000
2nd yr 47,000 x 4% = 48,880
3rd yr 48,880 x 5.6% = 51,617.28
4th yr 51,617.28 x 11.1% = 57,346.7981 or 57,346.80
