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Dennis_Churaev [7]
3 years ago
9

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

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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
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Step-by-step explanation:

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3 years ago
What are the factors of the quadratic function graphed below?
natulia [17]

Answer:

B

Step-by-step explanation:

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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
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