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2 years ago

Carmen's golf score is 6 strokes less than Linda's. Their two scores total 156. What is each girls score?

Mathematics

1 answer:

DochEvi [55]2 years ago

5 0

Answer:

Carmen's score is 75 strokes.

Linda's score is 81 strokes.

Step-by-step explanation:

Let the score of Linda be "x".

It is given that Carmen's golf score is 6 strokes less than Linda's.

Thus, Carmen's score is (x-6).

The total score of the girls is

= $x + (x-6) = 2x-6$

The total score is given to be 156.

Thus, the equation becomes,

$2x-6 = 156$

$2x= 162$

$x = 81$

Thus Linda's score is 81 and Carmen's score is (156-81) 75.

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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:

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

A local company makes snack-size bags of potato chips. Each day, the company produces batches of 400 snack-size bags using a pro

ioda

Answer:

Probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

Step-by-step explanation:

We are given that the company produces batches of 400 snack-size bags using a process designed to fill each bag with an average of 2 ounces of potato chips. Assume the amount placed in each of the 400 bags is normally distributed and has a standard deviation of 0.1 ounce.

Also, sample of 40 bags are selected.

Let $\bar X$ = sample average weight

The z-score probability distribution for sample mean is given by ;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean weight of potato chips = 2 ounces

$\sigma$ = standard deviation = 0.1 ounces

n = sample of bags = 40

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, the probability that a sample of 40 bags has an average weight of at least 2.02 ounces is given by = P( $\bar X$ $\geq$ 2.02 ounces)

P( $\bar X$ $\geq$ 2.02) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{2.02-2}{\frac{0.1}{\sqrt{40} } }$ ) = P(Z $\geq$ 1.265) = 1 - P(Z < 1.265)

= 1 - 0.89707 = 0.103

The above probability is calculated using z table by looking at value of x = 1.265 which will lie between x = 1.26 and x = 1.27 in the z table which have an area of 0.89707.

Therefore, probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

3 0

2 years ago

A car wash charges two prices depending on the size of the vehicle. The table shows the number of cars and the number of SUVs wa

Scrat [10]

Answer:

see below

Step-by-step explanation:

In the system of equations, x represents the, price to wash a car and y represents the price to wash an SUV.

If the system of equations is graphed in a coordinate plane, the coordinates (x, y) of the intersection of the two lines are the price of the car wash (x) and the price of the SUV wash (y)______.

x=12 and y = 18

x+y = 12+18 = 30

The total cost of 1 car and 1 SUV is $_30____.

5x+4y = 5(12)+4(18) = 60+ 72 =132

The total earnings for hour 3 are $_132_____.

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2 years ago

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slava [35]

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