Quadrilateral A'B'C'D'A

B

C

D

A, prime, B, prime, C, prime, D, prime is the result of dilating quadrilateral ABCDABCDA, B, C, D about point AAA by a scale factor of \dfrac{1}{2}
2
1

start fraction, 1, divided by, 2, end fraction.

Quadrilateral A'B'C'D'A B C D A, Prime, B, Prime, C, Prime, D, Prime Is The Result Of Dilating Quadrilateral

Answers

Answer 1

Answer:

[tex]A' = (2,0)[/tex]

Step-by-step explanation:

Given

See attachment for ABCD

[tex]k = \frac{1}{2}[/tex] --- the scale factor

Required

The coordinates of A'

From the attachment, we have:

[tex]A = (4,0)[/tex]

So:

[tex]A' = k * A[/tex]

[tex]A' = \frac{1}{2} * (4,0)[/tex]

[tex]A' = (2,0)[/tex]

Answer 2

Answer:

on khan, both are false

Step-by-step explanation:


Related Questions

what is the formula for perimeter of a square​

Answers

Answer: P = 4s

Step-by-step explanation:

P = 4s where s = the length of each side.  

Since each side of a square is the same length, the side length is multiplied by 4.

For the problem I tried dividing but my answers were not correct. How can I solve this problem then? Can someone help me out here please?

Answers

Answer:

8

Step-by-step explanation:

5 = 40

1 = x

Then we multiply by the rule of crisscrossing

5 x X = 40 x 1

5x = 40 then divide both by 5

X = 8

I can’t remember how to solve this?

Answers

Answer:

Step-by-step explanation:

[tex]\frac{(5.27+x)}{2} =-4.51[/tex],[tex]\frac{8.21+y}{2} = 1.37[/tex]

(3.75,-5.47)

Can someone please help me with my hw 20 points?

Answers

Answer:

The first equation; x-2y=8

Step-by-step explanation:

Hi there!

We're told that Ty wants to isolate x in one of the equations. To do so in either, he will need to use inverse operations to cancel out values and leave just x remaining on one side of the equation.

In the second equation, he would need to subtract both sides by 6y and then divide both sides by 4 to isolate x. It's a two-step process.

However, in the first equation, he only needs to add 2y to both sides to isolate x.

I hope this helps!

Answer:

using the first equation

cause Being that the first equation has the simplest coefficients (1, -2, for x, and y respectively), it seems logical to use it to develop a definition of one variable in terms of the other

suppose you have a bank account earning 6% annual interest rate compounded monthly, and you want to put in enough money so that you can withdraw $100 at the end of each month over a time frame of ten years. calculate how much money you need to start with. show work.

Answers

Answer:

maybe 10000

Step-by-step explanation:

Answer:

9007.35

Step-by-step explanation:

First find the effective rate: .06/12= .005

let x= amount

[tex]x=100\frac{1-(1+.005)^{-12*10}}{.005}\\100*\frac{1-.549632733}{.005}\\9007.345333[/tex]

PLESE HELp ANYONE. SOLVE ABC. ROUND YOUR ANSWERS TO THE NEAREST HUNDREDTH IF NECESSARY

Answers

Answer:

C=25°

a=11

b=12

Step-by-step explanation:

Find angle c,since angles in a triangle add up to 180 and we know angleA andB angle C will be

65+90+C=180

C=180-155

C=25°

To find a

use trig ratios

tanA=opposite/adjacent

tan65=a/5

a=tan65×5

a=10.72 round off to 11

To find b

sinC=opposite/hypotenuse

sin25=5/b

sin25 b=5

b=11.8 or rather 12

Answer:

Step-by-step explanation:

First find  side a and to find this  calculate tan 65

Tan 65 = [tex]\frac{opposite \ side}{adjacent\ side}=\frac{a}{5}\\\\[/tex]

2.144 = a/5

a = 2.144 * 5

b² = a² + c²

   = 121+25

   = 146.

b = √146 = 12.08 = 12

a = 10.72 = 11

Now find Tan C

[tex]Tan \ C = \frac{5}{10.72}\\\\Tan \ C = 0.4664\\[/tex]

C = tan⁻¹ 0.4664

C = 25°


Find the mean or average of these savings accounts $215, $156,$318, $75, and $25​

Answers

Answer:

157.8

Step-by-step explanation:

Add them all up to get 789 and divide them by 5 as there are five numbers to get the answer:)

debbie will be attending a concert at grand ole opry in nashville, tennessee. if the average number of songs performed there in a 10 day period is 167. approximately how many songs are performed there in a years time

Answers

Given:

The average number of songs performed there in a 10 day period is 167.

To find:

The number of songs performed there in a year time.

Solution:

We have,

Number of songs performed in 10 days = 167

Number of songs performed in 1 day = [tex]\dfrac{167}{10}[/tex]

                                                              = [tex]1.67[/tex]

We know that 1 year is equal to 365 days. So,

Number of songs performed in 365 day = [tex]1.67\times 365[/tex]

Number of songs performed in 1 year     = [tex]609.55[/tex]

                                                                   [tex]\approx 610[/tex]

Therefore, the number of songs performed there in a year time is about 610.

Find the maximum and the minimum value of the following objective​ function, and the value of x and y at which they occur. The function F=2x+16y subject to 5x+3y≤37, 3x+5y≤35, x≥0, y≥0
The maximum value of the objective function is ___ when x=___ and y=___

Answers

Answer:

The maximum value of the objective function is 112 when x = 0 and y = 7.

Step-by-step explanation:

Given the constraints:

5x+3y≤37, 3x+5y≤35, x≥0, y≥0

Plotting the above constraints using geogebra online graphing tool, we get the solution to the constraints as:

A(0, 7), B(7.4, 0), C(5, 4) and D(0, 0)

The objective function is given as E =2x+16y, therefore:

At point A(0, 7):  E = 2(0) + 16(7) = 112

At point B(7.4, 0): E = 2(7.4) + 16(0) = 14.8

At point C(5, 4): E = 2(5) + 16(4) = 74

At point D(0, 0): E = 2(0) + 16(0) = 0

Therefore the maximum value of the objective function is at A(0, 7).

The maximum value of the objective function is 112 when x = 0 and y = 7.

6. Convert 3−i into polar form and hence evaluate
[tex] {(3 - i)}^{7} [/tex]


Answers

9514 1404 393

Answer:

  ≈ 1000√10∠-129.04464° = -1992 -2456i

Step-by-step explanation:

  3 -i = √(3³+(-1)²)∠arctan(-1/3) ≈ √10∠-18.4349°

Then (3-i)^7 = 10^(7/2)∠(7×-18.4349°) = 1000√10∠-129.04464°

  = 1000√10(cos(-129.04464°) +i·sin(-129.04464°))

  = -1992 -2456i

2/3 - 10/9and5/3 and 7/9

Answers

Step-by-step explanation:

always Pythagoras with the coordinate differences as sides and the distance the Hypotenuse.

c² = (2/3 - 5/3)² + (-10/9 - -7/9)² = (-3/3)² + (-10/9 + 7/9)² =

= (-1)² + (-3/9)² = 1 + (-1/3)² = 1 + 1/9 = 10/9

c = sqrt(10)/3

Answer:

Step-by-step explanation:

Point 1  ([tex]\frac{2}{3}[/tex] , [tex]\frac{-10}{9}[/tex])   in the form (x1,y1)

Point 2 ( [tex]\frac{5}{3}[/tex] , [tex]\frac{-7}{9}[/tex])  in the form (x2,y2)

use the distance formula

dist = sqrt[ (x2-x1)^2 + (y2-y1)^2 ]

dist = sqrt [ [tex]\frac{5}{3}[/tex] -[tex]\frac{2}{3}[/tex])^2 + (  [tex]\frac{-7}{9}[/tex] - ( [tex]\frac{-10}{9}[/tex] ) )^2 ]

dist = sqrt [ ([tex]\frac{3}{3}[/tex])^2 + ([tex]\frac{3}{9}[/tex])^2 ]

dist = sqrt [  1 + ([tex]\frac{1}{3}[/tex])^2 ]

dist = sqrt [  [tex]\frac{9}{9}[/tex] + [tex]\frac{1}{9}[/tex] ]

dist = [tex]\sqrt{\frac{10}{9} }[/tex]

dist = [tex]\sqrt{10}[/tex] *[tex]\sqrt{\frac{1}{9} }[/tex]

dist = [tex]\sqrt{10}[/tex]  * [tex]\frac{1}{3}[/tex]

dist = [tex]\frac{\sqrt{10} }{3}[/tex]

The lifespan, in years, of a certain computer is exponentially distributed. The probability that its lifespan exceeds four years is 0.30. Let f(x) represent the density function of the computer's lifespan, in years, for x>0. Determine an expression for f(x).

Answers

Answer:

The correct answer is "[tex]0.300993e^{-0.300993x}[/tex]".

Step-by-step explanation:

According to the question,

⇒ [tex]P(x>4)=0.3[/tex]

We know that,

⇒ [tex]P(X > x) = e^{(-\lambda\times x)}[/tex]

⇒     [tex]e^{(-\lambda\times 4)} = 0.3[/tex]

∵ [tex]\lambda = 0.300993[/tex]

Now,

⇒ [tex]f(x) = \lambda e^{-\lambda x}[/tex]

By putting the value, we get

           [tex]=0.300993e^{-0.300993x}[/tex]


The triangles are similar. If QR = 9, QP = 6, and TU = 19, find TS. Round to the nearest tenth.
A) 16
B) 12.7
C) 2.8
D) 28.5

Answers

Answer:

TS = 12.7

Step-by-step explanation:

From the question given above, the following data were obtained:

QR = 9

QP = 6

TU = 19

TS =?

Since the triangles are SIMILAR, then,

QR / TU = QP / TS

With the above equation, we can obtain the value of TS as follow:

QR = 9

QP = 6

TU = 19

TS =?

QR / TU = QP / TS

9 / 19 = 6 / TS

Cross multiply

9 × TS = 19 × 6

9 × TS = 114

Divide both side by 9

TS = 114 / 9

TS = 12.7

Change 9/3 to percentage

Answers

Answer:

300%

Step-by-step explanation:

because 9/3×100=900/3=300 so it is 300%

Answer:

300%

Step-by-step explanation:

9/3 * 100%

900%/3 = 300%

HELP ASAP I WILL GIVE BRAINLIST

Convert 7π OVER 4 radians to degrees. Which quadrant does this angle lie in?
What are the sine, cosine and tangent of the angle 7π over 4? Be sure to show and explain all work.

Answers

Answer:

7π/4 radians = 315°, Quadrant IV

sin(315°) = -√2/2

cos(315°) = √2/2

tan(315°) = -1

Step-by-step explanation:

Mark looked at the statistics for his favorite baseball player, Jose Bautista. Mark looked at seasons
when Bautista played 100 or more games and found that Bautista's probability of hitting a home run
in a game is 0.173
If Mark uses the normal approximation of the binomial distribution, what will be the variance of
the number of home runs Bautista is projected to hit in 100 games? Answer choices are rounded
to the tenths place.
O 0.8
O 14.3
0 3.8
O 17.3

Answers

17.3

#CARRYONLEARNING

[tex]{hope it helps}}[/tex]

For two consecutive numbers, five times the number that is less is 3 more than 4 times the greater number, What are the numbers


This is due on 7/1/2021 at 8AM PST. Someone please help?

Answers

5x-3=12 I hope this is right I tried my best I am sorry if you get it wrong

The following data were collected from a simple random sample from an infinite population.
13 15 14 16 12
The point estimate of the population standard deviation is _____.
a. 1.581
b. 2.500
c. 2.000
d. 1.414

Answers

Answer:

1.581

Step-by-step explanation:

Given the data:

13 15 14 16 12

The point estimate of the standard deviation will be :

√Σ(x - mean)²/n-1

Mean = Σx / n = 70 / 5 = 14

√[(13 - 14)² + (15 - 14)² + (14 - 14)² + (16 - 14)² + (12 - 14)² / (5 - 1)]

The point estimate of standard deviation is :

1.581

Given the data points below, compute the sum of squared errors for the regression equation
Y
=
2
+
3
X
.


X

0

3

7

10

Y

5

5

27

31

Answers

Answer:

The sum of squared errors for the regression equation is 62.

Step-by-step explanation:

The sum of squared errors can be computed as follows:

X            Y           Y* = 2 + 3X         Y - Y*           (Y - Y*)^2

0            5                  2                      3                    9

3            5                  11                     -6                  36

7            27               23                     4                    16

10          31                32                    -1                     1  

20         68               68                     0                   62

From the above, we have:

Error = Y -  Y*

Error^2 = (Y - Y*)^2

Sum of squared errors = Sum of Error^2 = Total of (Y - Y*)^2 = 62

Therefore, the sum of squared errors for the regression equation is 62.

Plss help!!! Image included

Answers

Answer:

(-5,2)

Step-by-step explanation:

A function is a relation where each y-value has a unique x-value. That means that x's can never repeat. Therefore, to solve find the ordered pair that does not have the same x-value as one of the points on the graph. The x-values currently represented on the graph are -4, -2, 2, 3. So, the only coordinate pair that does not repeat an x-value is the last option, (-5, 2).

what graph shows the solution to the equation below log3(x+2)=1

Answers

Answer:

The solution to the equation  log3(x+2)=1 is given by x=1

Step-by-step explanation:

We are given that

[tex]log_3(x+2)=1[/tex]

We have to find the graph which shows the  solution to the equation log3(x+2)=1.

[tex]log_3(x+2)=1[/tex]

[tex]x+2=3^1[/tex]

Using the formula

[tex]lnx=y\implies x=e^y[/tex]

[tex]x+2=3[/tex]

[tex]x=3-2[/tex]

[tex]x=1[/tex]

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of a.

f(x)= 7x e^x, a= 0

Answers

Hi there!

[tex]\large\boxed{p(x) = 7x + 7x^2 + \frac{7}{2}x^3 + \frac{7}{6}x^4}[/tex]

Recall a Taylor series centered at x = 0:

[tex]p(x) = f(0) + f'(0)(x) + \frac{f''(0)}{2}x^{2} + \frac{f'''(0)}{3!}x^{3} + ...+ \frac{f^n}{n!}x^n[/tex]

Begin by finding the derivatives and evaluate at x = 0:

f(0) = 7(0)e⁰ = 0

f'(x) = 7eˣ + 7xeˣ   f'(0) = 7e⁰ + 7(0)e⁰ = 7

f''(x) = 7eˣ + 7eˣ + 7xeˣ  f''(0) = 7(1) + 7(1) + 0 = 14

f'''(x) = 7eˣ + 7eˣ + 7eˣ + 7xeˣ    f'''(0) = 21

f⁴(x) = 7eˣ + 7eˣ + 7eˣ + 7eˣ + 7xeˣ   f⁴(0) = 28

Now that we calculated 4 non-zero terms, we can write the Taylor series:

[tex]p(x) = 0 + 7x + \frac{14}{2}x^2 + \frac{21}{3!}x^3 + \frac{28}{4!}x^4[/tex]

Simplify:

[tex]p(x) = 7x + 7x^2 + \frac{7}{2}x^3 + \frac{7}{6}x^4[/tex]

Which statement is true about quadrilateral ABCD with vertices A(2, 8), B(3, 11), C(4, 8), and D(3, 5)?

Answers

Answer:

The quadrilateral is a rhombus

Step-by-step explanation:

Given

[tex]A = (2, 8)[/tex]

[tex]B = (3, 11)[/tex]

[tex]C = (4, 8)[/tex]

[tex]D=(3, 5)[/tex]

Required

The true statement

Calculate slope (m) using

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Calculate distance using:

[tex]d= \sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}[/tex]

Calculate slope and distance AB

[tex]m_{AB} = \frac{11 - 8}{3 - 2}[/tex]

[tex]m_{AB} = \frac{3}{1}[/tex]

[tex]m_{AB} = 3[/tex] -- slope

[tex]d_{AB}= \sqrt{(3 - 2)^2 + (11 -8)^2}[/tex]

[tex]d_{AB}= \sqrt{10}[/tex] -- distance

Calculate slope and distance BC

[tex]m_{BC} = \frac{8 - 11}{4 - 3}[/tex]

[tex]m_{BC} = \frac{- 3}{1}[/tex]

[tex]m_{BC} = -3[/tex] -- slope

[tex]d_{BC} = \sqrt{(4-3)^2+(8-11)^2[/tex]

[tex]d_{BC} = \sqrt{10}[/tex] --- distance

Calculate slope CD

[tex]m_{CD} = \frac{5 - 8}{3 - 4}[/tex]

[tex]m_{CD} = \frac{- 3}{- 1}[/tex]

[tex]m_{CD} = 3[/tex] -- slope

[tex]d_{CD} = \sqrt{(3-4)^2+(5-8)^2}[/tex]

[tex]d_{CD} = \sqrt{10}[/tex] -- distance

Calculate slope DA

[tex]m_{DA} = \frac{8 - 5}{2 - 3}[/tex]

[tex]m_{DA} = \frac{3}{- 1}[/tex]

[tex]m_{DA} = -3[/tex] -- slope

[tex]d_{DA} = \sqrt{(2-3)^2 + (8-5)^2}[/tex]

[tex]d_{DA} = \sqrt{10}[/tex]

From the computations above, we can see that all 4 sides are equal, i.e. [tex]\sqrt{10}[/tex]

And the slope of adjacent sides are negative reciprocal, i.e.

[tex]m_{AB} = 3[/tex]  and [tex]m_{CD} = -3[/tex]

[tex]m_{CD} = 3[/tex] and [tex]m_{DA} = -3[/tex]

The quadrilateral is a rhombus

Which of the following are rational numbers?

Answers

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

{7, -5, (2/3), 5.79}

»»————- ★ ————-««  

Here’s why:  

Rational numbers are numbers that could be written as a fraction with two integers.

⸻⸻⸻⸻

[tex]\boxed{\text{\underline{\textbf{Some Examples of Rational Numbers Are...}}}}\\\\\rightarrow \text{Integers}\\\\\rightarrow \text{Perfect Squares}\\\\\rightarrow \text{Terminating Decimals}\\\\\rightarrow \text{Recurring Decimals}[/tex]

⸻⸻⸻⸻

7 and -5 are integers, so they are rational. [tex]\frac{2}{3}[/tex] is already a fraction with integers. It is rational.5.79 is a terminating decimal. It is rational.The number π is a famous irrational number. It does not terminate nor repeat. [tex]\sqrt{13}[/tex] is not a perfect square. It is irrational.[tex]\sqrt{-4}[/tex] is a perfect square, but it is simplified to a complex number. Complex numbers are not rational.

⸻⸻⸻⸻

The rational numbers are {7, -5, (2/3), 5.79}.

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

can someone help with this​

Answers

Answer:

[tex]\frac{8}{45}[/tex]

Step-by-step explanation:

'of' means 'multiply'

4/5 × 2/9 = 8/45

PLEASE HELP!!! WILL GIVE BRAINLIEST!!!!

Answers

Finding the line of best fit is something a Machine Learning Model would do.

This particular ML model is called "Linear Regressor" or "Linear Regression Model". Look it up and there are definitely calculators for it, as it is relatively simple.

You can also, if you know how to use ML libraries and code, use Python to determine the value of [tex]b[/tex].

Hope this helps.

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a standard deviation of 118. Determine the 26th percentile for the number of chocolate chips in a bag. ​(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags. ​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Answers

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that [tex]\mu = 1262, \sigma = 118[/tex]

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.643 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -0.643*118[/tex]

[tex]X = 1186[/tex]

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.96 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -1.96*118[/tex]

[tex]X = 1031[/tex]

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.96 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = 1.96*118[/tex]

[tex]X = 1493[/tex]

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -0.675*118[/tex]

[tex]X = 1182[/tex]

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.675 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = 0.675*118[/tex]

[tex]X = 1342[/tex]

IQR:

1342 - 1182 = 160

Solve by using matrices. 2x – y +2 + w = -3 x + 2y – 3z + w = 12 3x - y - + 2w = 3 -2x + 3y + 2 – 3w = -3​

Answers

Some symbols and numbers are missing. I assume the system is supposed to read

2x - y + 2z + w = -3

x + 2y - 3z + w = 12

3x - y - z + 2w = 3

-2x + 3y + 2z - 3w = -3

In matrix form, this is

[tex]\begin{bmatrix}2&-1&2&1\\1&2&-3&1\\3&-1&-1&2\\-2&3&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-3\\12\\3\-3\end{bmatrix}[/tex]

which we can strip down to the augmented matrix,

[tex]\left[\begin{array}{cccc|c}2&-1&2&1&-3\\1&2&-3&1&12\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]

Now for the row operations:

• swap rows 1 and 2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\2&-1&2&1&-3\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]

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

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&-7&8&-1&-33\\0&7&-4&-1&21\end{array}\right][/tex]

• add 7 (row 2) to -5 (row 3), and row 3 to row 4

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&16&-2&-24\\0&0&4&-2&-12\end{array}\right][/tex]

• multiply through rows 3 and 4 by 1/2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&8&-1&-12\\0&0&2&-1&-6\end{array}\right][/tex]

• add -4 (row 4) to row 3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&0&3&12\\0&0&2&-1&-6\end{array}\right][/tex]

• swap rows 3 and 4

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&3&12\end{array}\right][/tex]

• multiply through row 4 by 1/3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&1&4\end{array}\right][/tex]

• add row 4 to row 3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&0&-2\\0&0&0&1&4\end{array}\right][/tex]

• multiply through row 3 by 1/2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• add -8 (row 3) and row 4 to row 2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&0&0&-15\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• multiply through row 2 by -1/5

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• add -2 (row 2) and 3 (row 3) and -1 (row 4) to row 1

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

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

Two groups were moving from one campsite to another. The first group traveled the distance in 5 hours. The second group finished in 7 hours. Find the distance between the campsites if the first group was going 4mph faster than the second group.

Answers

Answer:

The distance between the campsites was 70 miles.

Step-by-step explanation:

Since two groups were moving from one campsite to another, and the first group traveled the distance in 5 hours while the second group finished in 7 hours, to find the distance between the campsites if the first group was going 4mph faster than the second group, the following calculation must be performed:

X + 4 = 5

X = 7

4 x 7 = 28

(4 + 4) x 5 = 40

10 x 7 = 70

(10 + 4) x 5 = 70

Therefore, the distance between the campsites was 70 miles.

The weights of certain machine components are normally distributed with a mean of 5.19 ounces and a standard deviation of 0.05 ounces. Find the two weights that separate the top 8% and the bottom 8%. These weights could serve as limits used to identify which components should be rejected

Answers

Answer:

The  weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.

Step-by-step explanation:

We are given that

Mean,[tex]\mu=5.19[/tex]

Standard deviation,[tex]\sigma=0.05[/tex]

We have to find the two weights that separate the top 8% and the bottom 8%.

Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.

Z-value for p-value =0.08 =[tex]-1.41[/tex]

For 8% bottom

[tex]Z=\frac{x_1-\mu}{\sigma}=-1.41[/tex]

[tex]\frac{x_1-5.19}{0.05}=-1.41[/tex]

[tex]x_1-5.19=-1.41\times 0.05[/tex]

[tex]x_1=-1.41\times 0.05+5.19[/tex]

[tex]x_1=5.1195[/tex]

For 8% top

p-Value=1-0.08=0.92

Z- value=1.41

Now,

[tex]\frac{x_2-5.19}{0.05}=1.41[/tex]

[tex]x_2-5.19=1.41\times 0.05[/tex]

[tex]x_2=1.41\times 0.05+5.19[/tex]

[tex]x_2=5.2605[/tex]

(x1,x2)=(5.1195,5.2605)

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