In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. a winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. the two cards can be in any order. find the probability of being dealt a blackjack hand. what approximate percentage of hands are winning blackjack hands?

We are given here that a blackjack hand consists of:

1 of the 4 aces = 4 / 52

1 of the 16 cards worth 10 points (10, jack, queen, king) = 16 / 52

So assuming that cards are dealt without replacement, therefore the probability of getting a blackjack hand is:

P = 1st is ace * 2nd is 10 pt card + 1st is 10 pt card * 2nd is ace

P = (4 / 52) * (16 / 51) + (16 / 52) * (4 / 51)

P = 0.04827 = 4.83%

Therefore there is a 4.83% probability to get a blackjack hand.

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).