Answer:
relation of FDs are {a, b, d, h}. From these all but {h}
seem plausibly part of keys. Calculating the closures of likely candidates gives:
{a, b} + = {a b, c, i}
{b, d} + = {b, d, e, f}
{a, d} + = {a, d, g, h, i, j}
{a, b, d} + = {a, b, c, d, e, f, g, h, i, j}
decompose r into 2nf and then 3nf relations shown in attachment.
Step- by-Step Explanation
Explanation is given in the attachment.
It's a linear function. We need only two points to draw a graph.
We choose any values of x and calculate the value of y.
for x= 0 → y = 3(0) - 1 = 0 - 1 = -1 → (0, -1)
for x = 2 → y = 3(2) - 1 = 6 - 1 = 5 → (2, 5)
for x = 0 → y = 3(0) - 1/3 = 0 - 1/3 = -1/3 → (0, -1/3)
for x = 2 → y = 3(2) - 1/3 = 6 - 1/3 = 5 2/3 → (2, 5 2/3)
The answer is 1/12
The chance of getting a 3 on a six-sided dice is 1/6
The possibility of getting heads on a two-sided coin is 1/2
6 multiplied by 2 is 12
1 multiplied by 1 is 1
Hope it helps!
