Find the probability of drawing a 10, replacing it, and drawing an ace.
1 answer:
Assuming that you're talking about a standard deck of 52 cards, there are 4 tens and 4 aces.
So, at the beginning, you have probability 4/52=1/13 to draw a ten.
You replace, and you have an identical probability to draw an ace (you still have 4 favourable cases over 52 total cases).
Multiply the two probabilities to find the answer:
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given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then
the x-coordinate of the vertex is
= -
y = x² - 6x + 4 is in standard form
with a = 1, b = - 6 and c = 4, hence
= -
= 3
substitute this value into the equation for y- coordinate
y = 3² - 6(3) + 4 = 9 - 18 + 4 = - 5
vertex = (3, - 5 ) → second table