All categories

3 years ago

Solve the system of linear equations

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Answer:

Infinitely many solutions.

Step-by-step explanation:

3x-3y=6

-6x+6y=-12

------------------

2(3x-3y)=2(6)

-6x+6y=-12

-------------------

6x-6y=12

-6x+6y=-12

--------------------

0=0

You might be interested in

What is the value of

vladimir1956 [14]

The answer is x = -19

Solution:

9(x + 1) = -162

Expand:

9x + 9 = -162
9x = -162 - 9
9x = -171
x = -171/9
x = -19

6 0

2 years ago

I need help on this question can some help

guapka [62]

Answer:There’s a that you can download with a ton of edginuity answers if you type down edginuity answers and hit the pdf by reservamesa and you can find some.

Step-by-step explanation:

5 0

3 years ago

What are the factors of the quadratic function graphed below?

natulia [17]

Answer:

Step-by-step explanation:

The zeros of the quadratic occur at x = 0 and x = 6

Thus the factors are x and x - 6

4 0

3 years ago

A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function

dexar [7]

Here's our equation.

$h=-16t+64t+3$

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

$0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}$

$= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}$

So the ball will return to the ground at the positive value of $\boxed{\frac{8+\sqrt{67}}{4}}$ seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

$\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2$

$h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}$

8 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Solve the system of linear equations ​

Solve the system of linear equations