Select all of the relations that are functions. {(a, 1), (b, 1), (c, 1)} {(a, a), (a, b), (a, c)} {(1, a), (2, a), (3, a)} {(a,
netineya [11]
Answer:
{(a, 1), (b, 1), (c, 1)}, {(1, a), (2, a), (3, a)}, and {(a, a), (b, b), (c, c)}
**All functions are relations, but not ll relations are functions**
In a function, the x values CAN NOT repeat, only y-values can.
Answer:
8
Step-by-step explanation:
3 times 4 is 12 and 2 times 4 is 8
The conditional probability illustrates that's there's a 2/8 that the event A occurs.
<h3>How to illustrate the probability?</h3>
It should be noted that probability simply means the likelihood of the occurence of an event.
In this case, it can be delivered that P(AID) and P(DIA) aren't equal.
Hence, P(D|A) has event A as its given event, resulting in 2/8 for a probability.
Learn more about probability on:
brainly.com/question/24756209
#SPJ1
Answer:
x = 0
, y = 4
Step-by-step explanation:
Solve the following system:
{y = 4 - 3 x | (equation 1)
x + 2 y = 8 | (equation 2)
Express the system in standard form:
{3 x + y = 4 | (equation 1)
x + 2 y = 8 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{3 x + y = 4 | (equation 1)
0 x+(5 y)/3 = 20/3 | (equation 2)
Multiply equation 2 by 3/5:
{3 x + y = 4 | (equation 1)
0 x+y = 4 | (equation 2)
Subtract equation 2 from equation 1:
{3 x+0 y = 0 | (equation 1)
0 x+y = 4 | (equation 2)
Divide equation 1 by 3:
{x+0 y = 0 | (equation 1)
0 x+y = 4 | (equation 2)
Collect results:
Answer: {x = 0
, y = 4
Answer:
I am in middle school so I actually don't know the answer
