Answer:
y = -319/20| x = 377/20
Step-by-step explanation:
/| 7/2-x-y-(3/5) = 0| -11*x-(13*y) = 0
We try to solve the equation: 7/2-x-y-(3/5) = 0
29/10-x-y = 0 // - 29/10-x
-y = -(29/10-x) // * -1
y = 29/10-x
We insert the solution into one of the initial equations of our system of equations
We get a system of equations:
/| -(13*(29/10-x))-11*x = 0| y = 29/10-x
-1*13*(29/10-x)-11*x = 0
2*x-377/10 = 0
2*x-377/10 = 0 // + 377/10
2*x = 377/10 // : 2
x = 377/10/2
x = 377/20
We insert the solution into one of the initial equations of our system of equations
For y = 29/10-x:
y = 29/10-377/20
y = -319/20
We get a system of equations:
/| y = -319/20| x = 377/20
Answer: 5 cents less
Step-by-step explanation:
The unit price for the 24 pack would be $1.10 and the unit price for the 12 pack would be $1.15.
Answer:
Step-by-step explanation:
90 by 20%
10%=9
20%=18
90-18=62
Hope u understand
We get tne n-th term with: tn=1/3tn-1
We know that the first term is t1=81, so 81=1/3t*1-1
81=1/3t-1
82=1/3t
82*3t=1
3t=1/82
t=1/82*3=1/246
The second term is: 1/
2*(1/246)*2)-1
We get the third term by replacing n with 3 and so on...
Multiples of 11 between 1 and 30:
11, 22
So there are 2 numbers that are multiples of 11 in the bin. There are a total of 30 cards, so the probability is written as 2/30. Or we can simplify it to 1/15.
For the next question:
There are a total of 3 + 8 = 11 balls in the bag.
The probability of choosing a red ball is 3/11.
The probability of choosing a green ball is 8/11.
Multiply the three fractions:
3/11 * 3/11 * 8/11 = 72/1331
So the probability is 72/1331.
For the last question:
A standard deck of cards has 52 cards.
There are 4 queens and 4 kings in the deck.
Probability of choosing a queen is 4/52, and the probability of choosing a king AFTER you already chose a queen is 4/51.
Multiply the two fractions:
4/52 * 4/51 = 16/2652
So the probability is 16/2652 or 4/663
