A standard American deck of cards contains 52 cards in four “suits”: ♣ clubs
♦ diamonds
♥ hearts
♠ spades
Each card also has a “rank”. There are 13 ranks: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king), A(ace). Every suit-rank combination occurs in the deck, so a typical card might be called “the 7 of clubs” or “the queen of hearts”. Suppose we shuffle a deck (meaning that we put the cards in random order), then we draw a card
off the top. What is the probability that this card has an even number on it? (Jacks, queens, kings, and aces do not have numbers on them.)

I need the answer quickly pls

5/13

Step-by-step explanation:

We can first start by finding the number of even numbers in one suit and then multiply by 4. The even numbers are:

2, 4, 6, 8, 10

There are 5, and since there are 4 suits, there is a total of 20 even cards.

Probability is equal to the amount of desirable outcomes over the total number of outcomes. The desirable outcomes are when we pull out an even number, and as we calculated earlier, there are 20 of those. There are 52 total cards, so that is the number of total outcomes. We get the expression:

20 / 52 which simplifies to 5/13

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this