Four students, Jackson, Chloe, Lydia, and Noah, line up one behind the other. How

Sati [7]

The solution
m = s + 5
m - 4 = (3s) - 3 
(s + 5) - 4 = 3s - 3 
s + 1 = 3s - 3 
s = 3s - 4 
-2s = -4 
s = 2 
m = 7 
Check this: 

m - 4 = 3s - 3 
7 - 4 = 3(2) - 3 
3 = 6 - 3 
3 = 3 

So Sandy is 2 and Megan is 7.

5 0

1 year ago

Julie knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if c

Mamont248 [21]

Answer:

-3.64

Step-by-step explanation:

6 0

1 year ago

HELP! 50 POINTS!

Alexxandr [17]

(4, -1) ---> x = 4 and y = -1

If you look at the graph you can se this

Mark points (0, 2) and (8, -4) in the coordinate plane

Plot the line going through both of these points.

Read the coordinates of the intersection of the line so you can see the solution!

6 0

1 year ago

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

Ivahew [28]

Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

Reliability of the Robot with backup of component 1 can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

= 1 - 0.98

= 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

Similarly, to find out the reliability of component 2:

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

Reliability of the Robot with backup of component 3 can be computed as:

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

Reliability of the Robot with backup of component 4 can be computed as:

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The chances of failure of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

So, the reliability for each of the component & its backup is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

The reliability of the robot with backups for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

4 0

1 year ago

(-2x²y)³ / (xy²z)² * (-yz)² Hello guys, im having trouble with this babe...

Masja [62]

$\frac{(-2x^2y)^3}{(xy^2z)^2(-yz)^2}$

On top: (-2)³ = -2 × -2 × -2 = -8
(x²)³ = x^6 (the exponents multiply)
and of course, (y)³ = y³

On the bottom: (xy²z)² = x² y^4 z²
(-yz)² = y²z²
Multiplying these together, the exponents add and we get x² y^6 z^4.

$\frac{-8x^6y^3}{x^2y^6z^4}$

So, your reasoning is correct for what you have so far.

Your next step would be cancelling shared factors from the top and bottom.
Just like with regular fractions, if the numerator and denominator are divisible by the same number, you can divide them by it to simplify. (ex: 4/6 = 2/3)

Well, x^6 and x^2 are both divisible by x^2, right?
We can also cancel the y^3.

It might help to visualise the factors like this:

$\frac{-8xxxxxxyyy}{xxyyyyyyzzzz}$

Once you've cancelled out x² and y³ from each, you're left with

$\boxed{\frac{-8x^4}{y^3z^4}}$

7 0

2 years ago