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yawa3891 [41]
3 years ago
10

Question is in picture.

Mathematics
1 answer:
Norma-Jean [14]3 years ago
3 0

Answer:

<h2>B. 2x + y = 4</h2>

Step-by-step explanation:

Having the system of equations in its simplest form

\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right

If

a=d,\ b=e,\ c=f

then the system of equations has infinitely many solutions.

If

a=d,\ b=e,\ c\neq f

then the system of equations has no solution.

If

a\neq d\ or\ b\neq e

then the system of equations has one solution.

We have the equation

y=-2x+4

Convert to the standard form Ax + By = C<em>:</em>

<em />y=-2x+4<em>              add 2x to both sides</em>

y+2x=-2x+2x+4\\\\2x+y=4

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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
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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

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3 0
2 years ago
Read 2 more answers
Match the values associated with this data set to their correct descriptions. {6, 47, 49, 15, 43, 41, 7, 36}
Brums [2.3K]

1st quartile: 11

median: 38.50000

3rd quartile: 45

<h3>According to the given information:</h3>
  • Order these numbers in increasing order: 6, 7, 15, 36, 41, 43, 47, 49
  • There is a 38.5 median (it is the mean of 36 and 41 - the pair of middle entries).
  • 6,7,15,36, or the left-most half of the data, make up the sample.
  • The median of the lower half is 11, which is the first quartile (it is the mean of 7 and 15 - the pair of middle entries).
  • 41, 43, 47, and 49, which are the data points in the upper half, are to the right of the median.
  • The median of the upper half is 45 in the third quartile (it is the mean of 43 and 47 - the pair of middle entries).
  • The biggest value deviates 10.5 from the median (49-38.5)

Measure descriptive statistics

1st quartile: 11

median: 38.50000

3rd quartile: 45

To know more about quartile visit:

brainly.com/question/8737601

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I understand that the question you are looking for is :

2 Drag the tiles to the boxes to form correct pairs. Match the values associated with this data set to their correct descriptions.  {6, 47, 49, 15, 43, 41, 7, 36} first quartile 38.5  median 11  third quartile 10.5 the difference of the largest value and the median 45

7 0
2 years ago
A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ
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Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = 2.2\times 7.4=16.28

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = 2.2\times -6.2=-13.64

Thus change in moment is: 16.28-(-13.64)=29.92

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

\frac{29.92}{0.25}=119.68

Hence, the constant force exerted on the ball by the wall is 119.68 N.

6 0
3 years ago
The perimeter of a picture frame is 68 inches. The difference between the length of the frame and th
Julli [10]

Answer:

L - W = 2 and 2L + 2W = 68

Step-by-step explanation:

The two equations here are going to be-

Length minus width equals 2

L - W = 2

And

Length x 2 + Width x 2 = 68

2L + 2W = 68

6 0
2 years ago
What is the cost of 2 kilograms of flour if 3 kilograms cost $4.86 and the unit price for each packages of flour is the same
Pavlova-9 [17]
The answer is: $3.24.

\frac{4.86}{3}  =  \frac{x}{2}
\frac{9 .72}{3}  =  \frac{3x}{3}
3.24 = x
so ye...
8 0
3 years ago
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