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2 years ago

I’m not sure how to do this can anyone help!!!!!

Mathematics

1 answer:

satela [25.4K]2 years ago

7 0

9514 1404 393

Answer:

22/44

Step-by-step explanation:

The probability of black is the ratio of black cards to all cards. If the Ace and 2 are removed from each suit, there will be 11 of the 13 cards remaining. 2 suits are black, for a total of 2×11 = 22 black cards. Of course there are 4 suits altogether, for a total of 4×11 = 44 total cards. Then you have ...

P(black) = (black cards)/(total cards)

P(black) = 22/44

_____

You are expected to be familiar with the fact that a deck of 52 playing cards consists of 4 suits: diamonds, clubs, hearts, spades. Each suit has 13 cards, identified as Ace, 2, 3, ..., 10, Jack, Queen, King. The clubs and spades are black; the diamonds and hearts are red.

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irinina [24]

The probability of drawing a two and another 2 is 1/72

<h3>What is probability?</h3>

Probability is the likelihood or chance that an event will occur

Since the total number in the board game is 9, hence the probability of drawing a 2 will be 1/9

The probability of drawing the second 2 will be 1/8

Pr(drawing a 2, then drawing another 2) = 1/9 * 1/8

Pr(drawing a 2, then drawing another 2) = 1/72

Hence the probability of drawing a two and another 2 is 1/72

learn more on probability here: brainly.com/question/24756209

4 0

1 year ago

1. In an auditorium, there are 21 seats in the first row and 26 seats in the second row. The number of seats in a row continues

kondor19780726 [428]

1. Let $s_n$ be the number of seats in the $n$ -th row. The number seats in the $n$ -th row relative to the number of seats in the $(n-1)$ -th row is given by the recursive rule

$s_n=s_{n-1}+5$

Since $s_1=21$ , we have

$s_2=s_1+5$
$s_3=s_2+5=s_1+2\cdot5$
$s_4=s_3+5=s_1+3\cdot5$
$\cdots$
$s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5$

So the explicit rule for the sequence $s_n$ is

$s_n=21+5(n-1)\implies s_n=5n+16$

In the 15th row, the number of seats is

$s_{15}=5(15)+16=91$

2. Let $p_n$ be the amount of profit in the $n$ -th year. If the profits increase by 6% each year, we would have

$p_2=p_1+0.06p_1=1.06p_1$
$p_3=1.06p_2=1.06^2p_1$
$p_4=1.06p_3=1.06^3p_1$
$\cdots$
$p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1$

with $p_1=40,000$ .

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of

$p_{20}=1.06^{19}\cdot40,000\approx121,024$

3. Now let $p_n$ denote the number of pushups done in the $n$ -th week. Since $3\cdot4=12$ , $12\cdot4=48$ , and $48\cdot4=192$ , it looks like we can expect the number of pushups to quadruple per week. So,

$p_n=4p_{n-1}$

starting with $p_1=3$ .

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

$p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3$

6 0

3 years ago

PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

8 0

3 years ago

What is 42x - 7y ? Find the answer to the question please.

zvonat [6]

<span>42x - 7y = 7(6x - y)

</span>
<span>I hope this helps</span>

6 0

3 years ago

Read 2 more answers

Help with number 6. <br> A through F

Elodia [21]

A) always

Example
-3+(-3)= -9

B) never

C) never
D) always
E)never
F)sometimes

3 0

2 years ago