For the following system of equations, what is the x-value of the solution? -x+2=6 6y=x+18
2 answers:
Answer:
Option B is correct.
The value of x is, 0
Step-by-step explanation:
Given the system of equation:
-x+2y =6 .....[1]
6y =x+18 ......[2]
We can write equation [1] as;
-x = 6-2y or
x = -6 + 2y ......[3]
Substitute equation [3] in [2] we get;
6y =-6 + 2y + 18
or
6y = 2y +12
Subtract 2y from both sides we get;
6y -2y = 2y+ 12 -2y
Simplify:
4y = 12
Divide both sides by 4 we get;
Simplify:
y =3
Substitute the value of y in equation [3] to solve for x;
x = -6 + 2(3)
x = -6 +6 = 0
Therefore, the value of x is, 0
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The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>