All categories

3 years ago

Which number is prime 57, 63, 51, 67

Mathematics

2 answers:

AlladinOne [14]3 years ago

8 0

Prime numbers cannot be broken down anymore, and are only have factors of it's own number and 1.

67 is your answer, as it can only be factored down to 67 x 1.

hope this helps

gavmur [86]3 years ago

4 0

The best answer would be D) 67 because the other numbers are not prime.

You might be interested in

Brainliest !!!! Multiple choice

prisoha [69]

Answer:

Step-by-step explanation:

4 0

3 years ago

What is the equation of the line in standard form?

belka [17]

Answer:

$y = 6x + 4 \\ slop = 6 \\ y - y0 = slop(x - x0) \\ y - ( - 8) = 6 \times (x - 1) \\ y + 8 = 6x - 6 \\ y = 6x - 6 - 8 = 6x - 14$

6 0

2 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 in $R_A$ . There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in $R_C$ and $R_D$ . (Either the 1 comes first in $R_C$ 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

In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police

Lyrx [107]

Answer:

Option 3 - 4 miles

Step-by-step explanation:

Given : In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. The distance between the library and the police station is 5 miles.

To find : How far apart are the police station and the fire station?

Solution :

Let the distance between police station and fire station be 'x'.

The distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station.

i.e. $2x-3$