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Amanda [17]
3 years ago
14

How many solutions does the system of equation have?

Mathematics
2 answers:
serious [3.7K]3 years ago
8 0
We can determine the number of solutions from the slope intercept form of the equations.

Slope intercept form of equation 1 is:

y=- \frac{1}{4}x+3/4

Slope intercept form of second equation is:

y=- \frac{1}{4}x+ \frac{9}{12} \\  \\ 
y=  - \frac{1}{4}x+ \frac{3}{4}

The slope and the y-intercept of both equations are the same. This means, the two lines are lying over each other and hence they infinite number of solutions.

So, the correct answer is option C
miss Akunina [59]3 years ago
6 0
The equation no solution. This is because the lines are parallel.
The general equaltion of a line is,

Y=mX+c, where m is the gradient and c is the Y-intercept.

For the first equation,
<span>4x + 16y = 12
16y=-4x+12
y=(-1/4)x+12/16
gradient =-1/4</span>

For the second equation, Y = (-1/4)x +9/12,  the gradient = -1/4
 If the gradients are equation then, the lines do not meet hence no solutions.

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Laura has 24 stamps from each of 6 different countries. She can fit 4 stamps on each display sheets of an album. How many displa
Ket [755]
B. 36
To find the number of sheets that Laura is going to fill with her stamps, we first have to find the total number of stamps.
If she has 24 stamps EACH from 6 countries, to get the total, we will multiply
24 x 6 = 144 total number of stamps.

to see how many sheets she will fill we will DIVIDE the total number of stamps by 4 because that is the number she can fit onto One display sheet.
144 ÷ 4 = 36
36 sheets will be used to put four stamps each from a total of 144 stamps.
3 0
3 years ago
PLZZ HELP ME THIS IS A NO RETAKE QUIZ IT WOULD BE GREATLY APPRECIATED
MariettaO [177]

Answer:122

Step-by-step explanation:

4 0
3 years ago
A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the
olchik [2.2K]

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

                    F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P (  3 ≤X≤ 5 ) we will set the limits.

                   F(X) = P ( 3=

- The Expected Value can be determined by sum to infinity of CDF:

                   E(X) = Σ ( 1 - F(X) )

                   E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\=  \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\=  \frac{1}{(n)} - \frac{1}{(n+1)}\\\\=  \frac{1}{(n+1)} - \frac{1}{(n+ 2)}

                   E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]  

                   E(X) = 1

8 0
3 years ago
PLEASE ANSWERRRRRRRRRR
mash [69]

Answer: OPTION B.

Step-by-step explanation:

Given the following System of equations:

\left \{ {{3x-5y=19} \atop {x+y=1}} \right.

You can use the Elimination Method to solve it. The steps are:

1. You can mutliply the second equation by -3.

2. Then you must add the equations.

3. Solve for the variable "y".

Then:

\left \{ {{3x-5y=19} \atop {-3x-3y=-3}} \right.\\.....................\\-8y=16\\\\y=\frac{16}{-8}\\\\y=-2

4. Now that you know the value of the variable "y", you must substitute it into any original equation.

5. The final step is to solve for "x" in order to find its value.

Then:

x+(-2)=1\\\\x=1+2\\\\x=3

Therefore, the solution is:

(3,-2)

8 0
3 years ago
Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math.  The LCD is sin(x) cos(x).  Multiplying that in to each term looks like this:

[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

\frac{1}{sin(x)cos(x)}

We could separate that fraction into 2:

\frac{1}{sin(x)}×\frac{1}{cos(x)}

\frac{1}{sin(x)}=csc(x)  and  \frac{1}{cos(x)}=sec(x)

Therefore, the simplification is

sec(x)csc(x)

5 0
3 years ago
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