11.p=-8, 17-8=9
12.y=11, 3*11=33+16=46
13.t=4, 4*4=16-14=2
14.x=9, -9x 9*8=82-9=72-9=62
15.z=4, 12*4=48-18=30
16.g=0, 4*0=0, 7+0=7
17.x=4, 9*4=36-24=-3
18.q=3, 18*3=48+2=50
19.c=2, 3*2=6-4.5=4.1
20.y=4, 9+4=13+4.8=17.4

8 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Easy geometry. help please

olchik [2.2K]

(x, y) --> (x + 5, y - 1)

A(3, -1) --> A'(3 + 5, -1 - 1) --> A'(8, -2)

B(6, 1) --> B'(6 + 5, 1 -1) --> B'(11, 0)

C(2, 4) --> C'(2 + 5, 4 - 1) --> C'(7, 3)

D(-1, 3) --> D'(-1 + 5, 3 - 1) --> D'(4, 2)

7 0

3 years ago

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