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love history [14]
2 years ago
3

Suppose you ride your bike uphill to work at a speed of at least x mph. Going downhill on the way home, your speed is 5 mph fast

er. Your average speed to and from work is at least 12 mph. Which is the correct inequality to model the situation?
Mathematics
2 answers:
elena-s [515]2 years ago
7 0

Answer:

c

Step-by-step explanation:

Lelechka [254]2 years ago
3 0
The answer would be 4 mph. 

Let x = rate on the return trip Then, x+8 = rate in the other direction Distance = (Rate)(Time), so Time = (Distance/(Rate) Time to campus + Time from campus = 7/3 So, 7/(x+8) + 7/x = 7/3 Divide both sides by 7:  1/(x+8) + 1/x = 1/3Multiply both sides by the LCD, 3x(x+8), to get3x + 3(x+8) = x(x+8)Simplifying, we have 6x + 24 = x2 + 8x now just combine like terms and simplify. 
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The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L = 10 lo
liubo4ka [24]

Answer: the answer is A

Brandon’s sound intensity is 1/4 the level of Ahmad’s mower

Step-by-step explanation:

6 0
2 years ago
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It rounded to the nearest dime what is the greatest amount of money that rounds to $105.40
Vsevolod [243]
The greatest amount of money that could be rounded to $105.40 would be $105.44. Any higher, and it would be rounded to $105.50.
8 0
2 years ago
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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:

<u>Component Reliabilities:</u>

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:

<u>Reliability of the Robot with backup of component 1</u> 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

<u>Similarly, to find out the reliability of component 2:</u>

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

<u>Reliability of the Robot with backup of component 3 can be computed as:</u>

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

<u>Reliability of the Robot with backup of component 4 can be computed as:</u>

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 <u>highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664</u>. So, <u>Component 4 should get the backup in order to achieve the highest reliability.</u>

<u></u>

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 <u>chances of failure</u> 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 <u>reliability for each of the component & its backup</u> 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

<u>The reliability of the robot with backups</u> 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

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

4 0
2 years ago
N JKL, mJ = 90, mK = 30, and mL = 60. Which of the following statements about JKL are true?
Leya [2.2K]
KL=2*JL
I tried to find the rest of the question online, and I'm not sure if radicals couldn't show up or something, but attached is an image of the rules for a 30-60-90 triangle, so hopefully this helps.

3 0
2 years ago
Christopher is analyzing a circle, y2 + x2 = 121, and a linear function g(x). Will they intersect?
serg [7]
The complete question and the picture in the attached figure

we have that
y^2 + x^2 = 121
is a circle with center at (0,0) and radius r=11 units
and
 g(x)
points (-1,14)  (0,12)  (1,10)

using a graph tool
see the attached figure

the answer is
A. Yes, at the positive x coordinates. 

7 0
2 years ago
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