What are the answer choices
Answer:
the solution of the system is:
x = 1 and y = 2.
Step-by-step explanation:
I suppose that we want to solve the equation:
-6*x + 6*y = 6
6*x + 3*y = 12
To solve this, we first need to isolate one of the variables in one of the equations.
Let's isolate y in the first equation:
6*y = 6 + 6*x
y = (6 + 6*x)/6
y = 6/6 + (6*x)/6
y = 1 + x
Now we can replace this in the other equation:
6*x + 3*(1 + x) = 12
6*x + 3 + 3*x = 12
9*x + 3 = 12
9*x = 12 - 3 = 9
x = 9/9 = 1
Now that we know that x = 1, we can replace this in the equation "y = 1 + x" to find the value of y.
y = 1 + (1) = 2
Then the solution of the system is:
x = 1 and y = 2.
Answer:
R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Step-by-step explanation:
Squaring both sides of equation:
I^2 = (ER)^2/(R^2 + (WL)^2)
<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)
<=>(ER)^2 - (IR)^2 = (IWL)^2
<=> R^2(E^2 - I^2) = (IWL)^2
<=> R^2 = (IWL)^2/(E^2 - I^2)
<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Hope this helps!
Answer:
Step-by-step explanation:
Total = 50
Males = 20
Females = 30
a) P(exactly 1 female) = P(1 female , 2 male)
(30C1 * 20C2) / (50C3) = 30*190/19,600 = 0.2908
b) P(at least 2 males) = 1 - (P(0 males) + P(1 males))
= 1 - 30C3 * 20C0/50C3 - 30C2 * 20C1/50C3
= 1 - 4060*1/19,600 - 435*20/19,600 = 6840/19600 = 0.3489
Answer:
178 ERASERS
Step-by-step explanation:
YOU NEED TO START BACKWARDS. HE BOUGHT 76 ERASERS AND HE HAS 193, SO WHATEVER HE HAD AT THE END OF TUESDAY+76 IS HOW MUCH HE HAS ON WEDNESDAY. 193-76=117. ON TUESDAY, HE STARTED WITH ERASERS AT END OF TUESDAY+39 ERASERS. 117+39=156. ON MONDAY, HE STARTED WITH END OF MONDAY+22 ERASERS. 156+22=178
