Pls help, and do all of the qs! Will do brainliest!!

melomori [17]

The values of x, y, and z are 1, 2, and 3 respectively after using the substitution method.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have a linear equation in three variable:

2x + 3y - z = 5 ...(1)

4x - y - z = -1 ...(2)

x + 4y + z = 12 ...(3)

$\rm \mathrm{Substitute\:}x=\dfrac{5-3y+z}{2}$ (from the equation 1 and 2)

After solving:

$\rm \begin{bmatrix}-7y+z+10=-1\\ \dfrac{5y+3z+5}{2}=12\end{bmatrix}$

From the above, two equations:
y =2

z =3

Plug the above values in equation 1

x = 1

Thus, the values of x, y, and z are 1, 2, and 3 respectively after using the substitution method.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

5 0

2 years ago

The school gym can hold 650 people. A permanent bleacher in the gymn holds 136 people. The event organizers are setting up 25 ro

igomit [66]

650 -136=514
514 divided by 25 = 20.56
So it would be 21
This will give you a little extra chairs but they have to be equal in each row

3 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

34x−5=4 x = <br> im confused

IRINA_888 [86]

Answer:

9/34

Step-by-step explanation:

Let us move all the numerical values to one side:

34x= 4+5

34x = 9

x=9/34

<em>I hope this helps! :)</em>

8 0

3 years ago

WILL PICK BRAINLIEST PLZ ANSWER

Mars2501 [29]

Answer:

Archaea

Step-by-step explanation:

Archaea because it can survive in the diverse environments and is distinctly different from Bacteria which can also survive.

3 0

2 years ago

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