The probability of occurrence for the events A, B and C is; 1/4.
<h3>What is the probability of occurrence of.the described events?</h3>
For the first event A in which case, there's no odd number on the first two rolls, the possible events are; EEE and EEO. Consequently, the required probability is;
Event A = 2/8 = 1/4.
For the event B in which case, there's an even number on both the first and last rolls; the possible events are; EEE and EOE. Consequently, the required probability is;
Event B = 2/8 = 1/4.
For the event C in which case, there's an odd number on each of the first two rolls; the possible events are; OOO and OOE. Consequently, the required probability is;
Event C = 2/8 = 1/4.
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Answer:
Step-by-step explanation:
(x-1)²+(y-2)² =4 compare the given equation with the general one
(x-h)² +(y-k)² =r², where (h, k) are coordinates of the center and r is radius
so center is at ( 1, 2) and radius is 2
It is A. 0.05 represents one of his candies, and x represents how many of those, 0.10 represents the other candy while y represents how many of those candies. 1.00 is the total it should come to.
Answer:
D
Step-by-step explanation:
out of the choices
A) 2x-y=-2
3x+2y=-10
B) x+4y=3
2x+5y=3
C) 2x+3y=5
x-4y=-14
D) 3x+4y=15
x+y=5
only the system in D has a solution of x=5 and y=0
solve by substitution x+y=5 so x=5-y
plug into the other equation so 3(5-y)+4y=15
distribute 15-3y+4y=15 then minus 15 on both sides and combine like terms
so y=0 then put that value back into the other equation
so x+y=5 equals x+0=5 so x=5 this means the solution to the system of equations is the point (5,0)
Hope this helps! :)
I don't see a mistake. You solve in parentheses first, then you solve exponents. If the exponent is 0 in this context, the answer would be 0.
