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klio [65]
2 years ago
8

Help wit this please

Mathematics
1 answer:
kotegsom [21]2 years ago
5 0

Answer:

1: Out: 5, 8, 11, 14

2: Out: 2, 7, 10, 11

3: Out: 11  Rule: Add 4

4: Out: 6  Rule: Add 4

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Help please !!!! This is really hard
ankoles [38]

Answer:

Option -c  :5^ -12  .  2^ 40

Step-by-step explanation:

(  5^6 . 2^8    )( 5^-2  . 2^5 )

(  5^6 . 5^-2  )   ( 2^8  . 2^5 )

5^ -12  .  2^ 40

Option -c :  5^ -12  .  2^ 40

I hope im right!!  

4 0
2 years ago
an animal shelter has a ratio of dogs to cats that 3.2 if they are 30 cats at the shelter how many dogs are there
Anton [14]
I think it would be 90? Wait for other answers by smarter people than me haha.
8 0
3 years ago
One marble is randomly drawn and then replaced from a jar containing two white marbles and one
melisa1 [442]

Answer:

1/3.

Step-by-step explanation:

Prob(Drawing white marble) = 2/3

Prob(Drawing a black marble on second draw ) =  1/2

So:

Prob(Drawing a white then a red marble) = 2/3 * 1/2

= 2/6

= 1/3.

4 0
2 years ago
A town has a population of 6000 and grows at 3.5% every year. To the nearest year, how long will it be until the population will
kotykmax [81]

Answer:

5 years.

Step-by-step explanation:

Every year the population increases by a factor 103.5% or  1.035.

The equation relating population and time in years is:

P = 6000(1.035)^t  (This is an example of  exponential growth)

So we have:

7100 = 6000(1.035)^t

(1.035)^t = 7100/6000

(1.035)^t = 1.1833

t ln 1.035 = ln 1.1833

t = ln 1.1833/ln 1.035

= 4.89.

7 0
3 years ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

#SPJ4

3 0
2 years ago
Read 2 more answers
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