All categories

2 years ago

Solve the following systems of 3-variable linear equations.50 points

Mathematics

1 answer:

Natali [406]2 years ago

7 0

Answer:

x = 9/25

y = 7/25

z = 4/25

Step-by-step explanation:

2x + y = 1 .......(1)

3y + z = 1 ........(2)

x + 4z = 1 ........(3)

Elimination 1 and 2

2x + y = 1 | ×3 |

3y + z = 1 | ×1 |

6x + 3y = 3

3y + z = 1

___________--

6x - z = 2 .............. (4)

Elimination 3 and 4

x + 4z = 1 | ×6 |

6x - z = 2 | ×1 |

6x + 24z = 6

6x - z = 2

___________--

25z = 4

z = 4/25

Elimination 3 and 4

x + 4z = 1 | ×1 |

6x - z = 2 | ×4 |

x + 4z = 1

24x - 4z = 8

___________+

25x = 9

x = 9/25

Subsitution 1

2x + y = 1

2(9/25) + y = 1

18/25 + y = 1

y = 1 - 18/25

y = 25/25 - 18/25

y = 7/25

You might be interested in

What fraction equals 48% in it’s simplest form

dmitriy555 [2]

Answer:

12/25

Step-by-step explanation:

48% is equal to 48/100

This can be simplified by dividing both the numerator and denominator by 4.

48/4=12

100/4=25

8 0

2 years ago

Read 2 more answers

I really need help with this problem steps

melamori03 [73]

Solve for y by setting one equation =to X after that you can substitute the equation into the other one let's set the first one in X= FORM
X-7y=10
-2x+14y=-20

Add 7y to both sides of the first equation to let x stand alone
X=7y+10
Now you can substitute x on the second equation
-2 (7y+10)+14y=-20
Distribute
-14y-20+14y=-20
Add and simplify
Us cancel out =0
-20=-20
0=0
They are the same equations you can also divide the second equation by -2 which would make it look like this
X-7y=10
-2 (x-7y=10)
Let me know if I have answered your question

5 0

2 years ago

The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine

salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

b. P(1000<x<2000) = 0.2325

c. P(x<1000) = 0.6321

d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = $\int\limits^b_a {P(x)} \, dx$

Then, for the electronic component, probability will be:

P(a<x<b) = $\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx$

P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

P(a<x<b) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

0.1 = $1-e^\frac{-x}{1000}$

$-e^{\frac{-x}{1000} }=-0.9$

${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0

3 years ago

Please help me out!

melisa1 [442]

Answer:

don't under stand your question please

7 0

2 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

2 years ago

Solve the following systems of 3-variable linear equations.​50 points

Solve the following systems of 3-variable linear equations.50 points