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polet [3.4K]
4 years ago
14

This is Matrix for pre calc

Mathematics
1 answer:
gavmur [86]4 years ago
6 0

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

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Step-by-step explanation:

6/39_=

_6÷39/3

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What is the mean of the following set of numbers? 35, 40, 56, 52, 67
baherus [9]

Answer:

The mean is 50

Step-by-step explanation:

To find the mean, add up all the numbers and divide by the number of numbers

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The amount Lin's sister earns at her part-time job is proportional to the number
jarptica [38.1K]

the equation framed in the form of y=kx is  y = 9.6(x) and ,  x as a function of y x=\frac{1}{ 9.6}(y) .

<u>Step-by-step explanation:</u>

Here we have , The amount Lin's sister earns at her part-time job is proportional to the number  of hours she works. She earns $9.60 per hour. We need to find  an equation in the form y=kx  to describe this situation, where x represents the hours she works and y  represents the dollars she earns.Is y a function of x . Also , Write  an equation describing x as a function of y . Let's find out:

Here , Lin's sister earns $9.60 per hour . Let  x represents the hours she works and y  represents the dollars she earns . So , According to question following is the equation framed in the form of y=kx :

⇒ y = 9.6(x)

Yes, y is a function of x , as a straight line with a slope of 9.6

Now ,  x as a function of y :

⇒ \frac{y}{ 9.6}=(x)

⇒ x=\frac{1}{ 9.6}(y)

Therefore , the equation framed in the form of y=kx is  y = 9.6(x) and ,  x as a function of y x=\frac{1}{ 9.6}(y) .

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The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ou
Setler [38]

Answer:

Step-by-step explanation:

(a)

The distribution of X is Normal Distribution with mean = \mu =17 and Variance = \sigma^{2} = 16 \ i.e., X \sim N (17, 16),

(b)

The distribution of \bar{x} is Normal Distribution with mean = \mu =17 and Variance = \sigma^{2}/n = 16/15= 1.0667.i.e., \bar{x}\sim N(17,1.0667)

c)

To find P(15.5 < X < 18):

Case 1: For X from 15.5 to mid value:

Z = (15.5 - 17)/4 = - 0.375

Table of Area Under Standard Normal Curve gives area = 0.1480

Case 2: For X from mid value to 18:

Z = (18 - 17)/4 = 0.25

Table of Area Under Standard Normal Curve gives area = 0.0987

So,

P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467

So,

Answer is:

0.2467

(d)

SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}

= 1.0328

To find P(15.5 < \bar{x}< 18):

Case 1: For \bar{x} from 15.5 to mid value:

Z = (15.5 - 17)/1.0328 = - 1.4524

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 18:

Z = (18 - 17)/1.0328 = 0.9682

Table of Area Under Standard Normal Curve gives area = 0.3340

So,

P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605

So,

Answer is:

0.7605

(e)

Correct option:

No

because Population SD is provided.

(f)

(i)

Q1 is given by:

- 0.6745 = (\bar{x} - 17)/1.0328

So,

X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034

So,

Q1 = 16.3034

(ii)

Q3 is given by:

0.6745 = (\bar{x} - 17)/1.0328

So,

X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966

So,

Q3= 17.6966

(iii)

IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932

So

Answers are:

Q1 = 16.3034 ounces

Q3 = 17.6966 Ounces

IQR = 1.3932 Ounces

8 0
3 years ago
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