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34kurt
2 years ago
6

Please help! Will give brainlyest if right!! Lots of points.

Mathematics
2 answers:
tankabanditka [31]2 years ago
8 0

Answer:

Look

Step-by-step explanation:

luda_lava [24]2 years ago
3 0

Answer:

D

Step-by-step explanation:

I can explain if you need. Write in comments if you do need.

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Im not sure help!!!!!!!
kiruha [24]
The answer would be B
5 0
3 years ago
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3 cards are chosen at random from a standard 52-card deck. what is the probability that they form a pair? (a 3-card hand is a 'p
kozerog [31]

Hello!

First of all, we have a 4/52, or 1/13 chance of selecting a pair. Then, we select 1/3 cards 3 times, so we use this as our other fraction.

\frac{1}{13}\cdot \frac{1}{3}=\frac{1}{39}

Therefore, we have a 1/39 or about 3%.

I hope this helps!

8 0
3 years ago
Simplify the expression.<br> 3^-9 * 3^6 * 3^6<br><br> 3^3<br> 9^3<br> 27^3<br> 3^-324
kow [346]
3^3. if you're simplifying then you keep the base the smae but add the exponents since you are "multiplying" the bases

7 0
3 years ago
Exercise 10.10.2: Distributing a coin collection. About A man is distributing his coin collection with 35 coins to his five gran
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

Given that:

all coins are same;

The same implies that the number of the non-negative integral solution of the equation:

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

Thus, the number of the non-negative integral solution is:

^{(35+3-1)}C_{5-1} = ^{39}C_4

(b)

Here all coins are distinct.

So; the number of distribution appears to be an equal number of ways in arranging 35 different objects as well as 5 - 1 - 4 identical objects

i.e.

= \dfrac{(35+4)!}{4!}

= \dfrac{39!}{4!}

(c)

Here; provided that the coins are the same and each grandchild gets the same.

Then;

x_1+x_2+x_3+x_4+x_5 = 35

x_1 > 0 ;  \ \ \ x_1 \  \varepsilon  \ Z

x_1=x_2=x_3=x_4=x_5

5x_1 = 35\\\\ x_1= \dfrac{35}{5} \\ \\  x_1= 7

Thus, each child will get 7 coins

(d)

Here; we need to divide the 35 coins into 5 groups, this process will be followed by distributing the coin.

The number of ways to group them into 5 groups = \dfrac{35!}{(7!)^55!}

Now, distributing them, we have:

\mathbf{\dfrac{35!}{(7!)^55!}  \times 5!= \dfrac{35!}{(7!)^5}}

3 0
3 years ago
There are 13 dolls in a line. In how many different ways can they be arranged?
rodikova [14]

Answer:

13

Step-by-step explanation:

8 0
3 years ago
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