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2 years ago

The cost of 6 sandwiches and 4 drinks is $53. The cost of 4 sandwiches and 6 drinks is $47. How much does one sandwich cost?

Mathematics

1 answer:

melomori [17]2 years ago

8 0

In this question, we're trying to find cost of one sandwich.

To find this, we must make a systems of equation from the given information:

We would represent sandwiches as "s" and drinks as "d".

Systems of equations:

6s + 4d = 53

4s + 6d = 47

Solve for s:

6(6s + 4d = 53)

-4(4s + 6d = 47)

36s + 24d = 318

-16s - 24d = -188

--------------------------

20s = 130

Divide both sides by s

s = 6.50

This means that one sandwich costs $6.50

Answer:

$6.50

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First 100 units = 24p per unit Remaining units = 16p per unit Mrs. Watt checks her electricity bill. Here are her meter readings

Talja [164]

Answer: a) 146 units used b) the cost of the number of units used is 3136 p.

Step-by-step explanation:

Since we have given that

First 100 units = 24 p per unit

Remaining units = 16 p per unit

New reading = 7143 units

Old reading = 6997 units

Difference between the readings would be

$7143-6997\\\\=146$

So, there are 146 units used.

(B) Calculate the cost of the number of units used

There are 146 units

For the first 100 units = 24 p per unit

For the remaining i.e. 146-1400 = 46 units , cost = 16 p per unit

Total cost of the number of units used is given by

$24\times 100+46\times 16\\\\=2400+736\\\\=3136\ p$

Hence, a) 146 units used b) the cost of the number of units used is 3136 p.

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2 years ago

a square garden has an area of 225 square feet. how much fencing will a gardener need to buy in order to place fencing around th

IgorC [24]

The gardener will need to buy 60 feet of fencing. Because the square root of 225 if 15, and 15 times 4 is 60. So, 60 is the perimeter of the garden.

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2 years ago

John drove from station A to station B a distance of 224 miles. On his way back, he increased his speed by 10 mph. If the journe

meriva

Answer: 70 mph

Step-by-step explanation:

224/x =224/x+10 +24/60

Solve equation

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2 years 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

2 years ago

What is the equation of a line in slope intercept form that contains the points (-4,3) and (2,-6)

Nataly_w [17]

Answer:

y = (-3/2)x - 3

Step-by-step explanation:

The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

Given two points, we can calculate the slope by dividing the change in y (or the difference in the y-coordinates) by the change in x (or the difference in the x-coordinates). Our two points are (-4, 3) and (2, -6):

m = (3 - (-6)) / (-4 - 2) = 9 / (-6) = -3/2

So, we can update our equation:

y = (-3/2)x + b

The y-intercept is where the graph crosses the y-axis, or the y-value where x = 0. Let's plug in 3 for y and -4 for x:

3 = (-3/2) * (-4) + b

3 = 6 + b

b = -3

So, our y-intercept is -3.

Our slope-intercept form is thus:

y = (-3/2)x - 3

~ an aesthetics lover

7 0

2 years ago