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liq [111]
2 years ago
15

evan has a job selling magazine and newspaper subscriptions. he earns $23 for each magazine subscription and $54 for each newspa

per subscription that he sells. his goal is to make more than $642 per week. he expects to sell at least 10 subscriptions per week. if m represents the number of magazine subscriptions sold and n represents the number of newspaper subscriptions sold, which system of inequalities models the given situation?
Mathematics
1 answer:
SashulF [63]2 years ago
5 0
First we write the variables already defined:
 m = the number of magazine subscriptions sold
 n = the number of newspaper subscriptions sold
 We now write the system of inequations based on the following facts:
 "he earns $ 23 for each magazine subscription and $ 54 for each newspaper subscription that he sells. his goal is to make more than $ 642 per week"
 23m + 54n> 642
 "I have expectations to sell at least 10 subscriptions per week"
 m + n> = 10
 Answer: 
 A system of inequalities that models the given situation is: 
 23m + 54n> 642 
 m + n> = 10
You might be interested in
Estimate the sum of 196 and 482
k0ka [10]
The answer to this problem would be 700. if you need the sum that isnt estimated, its 678.
4 0
2 years ago
Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.
Ghella [55]

Answer:  The required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

Step-by-step explanation:  Given that Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Also, Heloise has written \frac{1}{4} as many articles as Mustafa has and Gia has written \frac{3}{2} as many articles as Mustafa has.

We are to write an inequality to determine the number of articles, m, Mustafa could have written for the school newspaper.  Also, to solve the inequality.

Since m denotes the number of articles that Mustafa could have written. Then, according to the given information, we have

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.

And the solution of the above inequality is as follows :

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22\\\\\\\Rightarrow \dfrac{4x+x+6x}{4}>22\\\\\\\Rightarrow 11x>88\\\\\Rightarrow x>\dfrac{88}{11}\\\\\Rightarrow x>8.

Thus, the required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

4 0
2 years ago
Read 2 more answers
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
Jeff and Kayla are finding the product of 178 times 56.They both are using the distributive property to find partial products.Lo
UNO [17]

Answer:

9968

Step-by-step explanation:

178×56=178×(50+6)

=178×50+178×6

=8900+(170+8)×6

=8900+170×6+8×6

=8900+1020+48

=9968

6 0
2 years ago
Enter the expression N0e−λt, where N0 is N-naught (an N with a subscript zero) and λ is the lowercase Greek letter lambda.
Natali5045456 [20]
So the question tells to express the expression in your problem where N0 is N-naught and the symbol represent the lower case Greek letter lambda. So the best answer or expression would be that the lambda is the wavelength of the expression. I hope you are satisfied with my answer 
7 0
2 years ago
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