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EleoNora
2 months ago
13

Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest

and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?
Mathematics
1 answer:
Leona [12.6K]2 months ago
8 0

Answer:

It is necessary to invest $5,000 in the 8% option

And, $15,000 in the 12% option

Step-by-step explanation:

Let’s begin by calculating the total expected return.

With an anticipated total return of 11%, the expected amount back is;

11/100 * 20,000 = $2,200

Let x represent the amount put into the 8% investment, and y represent the investment in the 12% option

From this, we have the equation;

x + y = 20,000 ••••••(i)

Now considering the interest portion;

The interest for the 8% portion is 8/100 * x

For the 12% part, it’s 12/100 * y

Thus, we express that as;

(8/100 * x) + (12/100 * y) = 2,200

Consolidating, we have;

8x/100 + 12y/100 = 2,200

Multiplying throughout by 100 gives us

8x + 12y = 220,000 ••••••(ii)

Next, we can solve this system of equations;

x + y = 20,000

8x + 12y = 220,000

Using i, we find x = 20,000 - y

Substituting into ii yields;

8(20,000-y) + 12y = 220,000

160,000-8y + 12y = 220,000

Combining yields 4y = 220,000 - 160,000

Thus, 4y = 60,000

So, y = 60,000/4

y = $15,000

From this, x can be calculated as follows:


x = 20,000 - y

x = 20,000 - 15,000

x = $5,000

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One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit
tester [12383]

Answer:

a) Robot Reliability = 0.7876

b1) Component 1: 0.8034

    Component 2: 0.8270

    Component 3: 0.8349

    Component 4: 0.8664

b2) To maximize overall reliability, Component 4 should be backed up.

c) To achieve the highest reliability of 0.8681, backup for Component 4 with a reliability of 0.92 should be implemented.

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) The reliability of the robot can be determined by calculating the reliabilities of the individual components that constitute the robot.

Robot Reliability = R1 x R2 x R3 x R4

                                      = 0.98 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.787626 ≅ 0.7876

b1) As only a single backup can be used at once, and its reliability matches that of the original, we evaluate each component's backup sequentially:

Robot Reliability with Component 1 backup is calculated by first assessing the failure probability of the component plus its backup:

Failure probability = 1 - R1

                      = 1 - 0.98

                      = 0.02

Combined failure probability for Component 1 and backup = 0.02 x 0.02 = 0.0004

Thus, reliability of combined Component 1 and backup (R1B) = 1 - 0.0004 = 0.9996

Robot Reliability = R1B x R2 x R3 x R4

                                         = 0.9996 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.8034

To determine reliability of Component 2:

Failure probability for Component 2 = 1 - 0.95 = 0.05

Combined failure probability of Component 2 and backup = 0.05 x 0.05 = 0.0025

Reliability of Component 2 with backup (R2B) = 1 - 0.0025 = 0.9975

Robot Reliability = R1 x R2B x R3 x R4

                = 0.98 x 0.9975 x 0.94 x 0.90

Robot Reliability = 0.8270

Robot Reliability with backup of Component 3 calculates as follows:

Failure probability for Component 3 = 1 - 0.94 = 0.06

Combined failure probability of Component 3 and backup = 0.06 x 0.06 = 0.0036

Reliability for Component 3 with backup (R3B) = 1 - 0.0036 = 0.9964

Robot Reliability = R1 x R2 x R3B x R4  

                = 0.98 x 0.95 x 0.9964 x 0.90

Robot Reliability = 0.8349

Robot Reliability with Component 4 backup calculates as:

Failure probability for Component 4 = 1 - 0.90 = 0.10

Combined failure probability of Component 4 and backup = 0.10 x 0.10 = 0.01

Reliability for Component 4 and backup (R4B) = 1 - 0.01 = 0.99

Robot Reliability = R1 x R2 x R3 x R4B

                                      = 0.98 x 0.95 x 0.94 x 0.99

Robot Reliability = 0.8664

b2) The best reliability is achieved with the backup of Component 4, yielding a value of 0.8664. Thus, Component 4 is the best candidate for backup to optimize reliability.

c) A reliability of 0.92 indicates a failure probability of = 1 - 0.92 = 0.08

We can compute the probability of failure for each component along with its backup:

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

Thus, the reliabilities for each component and its backup become:

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

Reliability of robot including backups for each of the components can be calculated as:

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

              = 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Backup for Component 1 = 0.8024

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

              = 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Backup for Component 2 = 0.8258

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

              = 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Backup for Component 3 = 0.8339

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

              = 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Backup for Component 4 = 0.8681

To maximize overall reliability, Component 4 should be backed up at a reliability of 0.92, achieving an overall reliability of 0.8681.

4 0
2 months ago
A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Leona [12618]
(a) The likelihood that all 5 eggs chosen are unspoiled is 0.0531. (b) The probability that 2 or fewer out of the 5 eggs are unspoiled is 0.3959. (c) The probability that more than 1 of the selected 5 eggs are unspoiled is 0.8747. Step-by-step explanation: The complete query is: A subpar carton of 18 eggs has 8 that are spoiled. An unsuspecting chef selects 5 eggs at random for his “Mega-Omelet Surprise.” Calculate the probability of receiving (a) exactly 5 unspoiled eggs, (b) 2 or fewer, and (c) more than 1 unspoiled egg. Define X = number of unspoiled eggs. In the faulty carton, 8 eggs are spoiled. The probability of selecting an unspoiled egg is independent of others. Provided that a chef randomly picks 5 eggs, the variable X follows a Binomial distribution with parameters n = 5 and p = 0.556. Success is defined as selecting an unspoiled egg. The probability mass function of X is as follows: (a) Calculate the probability of selecting all unspoiled eggs. Thus, this probability is found to be 0.0531. (b) For 2 or less unspoiled eggs, the probability is computed: P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2), resulting in a probability of 0.3959. (c) For more than 1 unspoiled egg: P (X > 1) = 1 - P (X ≤ 1), yields a final probability of 0.8747.
6 0
2 months ago
Sue played four games of golf for these games her modal score was 98 and her mean score was 100 her range of score was 10 what w
zzz [12365]

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Explanation:

We know that

Mean = 100

Mode = 98

Range = 10

And from the formula,

Range = Highest - Lowest.

Let’s set the highest observation as x and the lowest as y.

Thus, we have the equation x - y = 10 (equation 1).

The observations can be represented as:

x, 98, 98, y.

Using the mean formula yields:

Mean = \frac{\text{Sum of observation}}{\text{N.of observaton}}.

This means our second equation is:

x + y = 204.

By applying the elimination method to solve these linear equations, we find:

x = 97.

and

x+y=204\\y=204-x\\y=204-97\\y=107.

Therefore, the other two observations are 97 and 107.

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lawyer [12517]
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