Step-by-step explanation:
Answer:
m=4, b=1
Step-by-step explanation:
Answer:
-2 or -5
Step-by-step explanation:
( shown in photo below )
9514 1404 393
Answer:
Step-by-step explanation:
The general approach here is to choose a variable you want to eliminate, identify the coefficients of it, and use those to multiply the equations so adding the products will eliminate the variable.
Here, the coefficients of x are -1 and -3; the coefficients of y are 2 and 5. We often like to choose the set of coefficients that includes 1 or where the coefficients are related by an integer factor. These criteria suggest that we should use the coefficients -1 and -3. One of our multipliers must be the opposite of one of these coefficients. So, we choose for them to be -3 and +1.
That is, multiplying the first equation by -3 will make the x-coefficient be 3; multiplying the second equation by 1 will make the x-coefficient be -3. Adding these equations will then eliminate x terms.
(-3)(-x +2y = 12)
<u>+ </u><u>(1)</u><u>(-3x +5y =27)</u>
<u>gives</u> (3x -6y) +(-3x +5y) = -36 +27
which simplifies to the sum ...
0x -1y = -9
__
Dividing by the coefficient of y, we find y = 9. Substituting into the first equation, we have ...
-x +2(9) = 12)
x = 18-12 = 6
The solution is (x, y) = (6, 9).
Answer:
0.99945
Step-by-step explanation:
The probability that a bridge hand of 13 cards would contain at least one card that is ten or higher would be the inverse of the probability that all of the 13 cards would not contain any 10 or higher, in other words all of them are 9 or lower. From 2 to 9 we have 8 cards of 4 suits like that, so a total of 32 out of 52 cards
The probability of this to happen is
In the first draw, chance are 32/52
In the 2nd draw, chance are 31/51
In the 3rd draw, chance are 30/50
...
In the 13th draw, chance are 20/40
So the total probability is
So the probability that a randomly selected bridge hand is a Yarborough is
1 - 0.000547 = 0.99945