If $f(x)$ is a polynomial of degree 3, and $g(x)$ is a polynomial of degree 5, then what is the degree of polynomial $2f(x) + 4g

When two straight lines cross, the vertically opposite angles created are equal, and the remaining two angles match each other as well. Suppose one known angle is x; then the two adjacent angles can be calculated by subtracting twice x from 360 degrees and dividing the remainder by 2.

Thus, the table fills out as follows:

Row 1:

Given angle <GEF = 120°

Angle <FEM is adjacent to <GEF, so


Angle <MEH is vertically opposite to <GEF, making it equal to 120°

Angle <HEG is vertically opposite <FEM, so it equals 60°.


Row 2:

Given angle <MEH = 150°

Since <MEH is vertically opposite to <GEF, <GEF = 150°

Angle <FEM is adjacent to <GEF, thus


Angle <HEG is vertically opposite to <FEM, so <HEG = 30°.


Row 3:

Given angle <FEM = 25°

Angle <FEM is adjacent to <GEF, thus


Angle <MEH is vertically opposite to <GEF, equal to 155°

Angle <HEG is vertically opposite <FEM, so it equals 25°.


Row 4:

Given angle <HEG = 45°

Angle <HEG is adjacent to <GEF, so


Angle <FEM is vertically opposite to <HEG and thus equals 45°

Angle <MEH is vertically opposite to <GEF, hence <MEH = 135°.

Answer:

Step-by-step explanation:

Considering the equation

Sin(5x) = ½

5x = arcSin(½)

5x = 30°

Then,

The general formula for sin is

5θ = n180 + (-1)ⁿθ

Dividing throughout by 5

θ = n•36 + (-1)ⁿ30/5

θ = 36n + (-1)ⁿ6

The solution range is

0<θ<2π which means 0<θ<360

First solution

When n = 0

θ = 36n + (-1)ⁿθ

θ = 0×36 + (-1)^0×6

θ = 6°

When n = 1

θ = 36n + (-1)ⁿ6

θ = 36-6

θ = 30°

When n = 2

θ = 36n + (-1)ⁿ6

θ = 36×2 + 6

θ = 78°

When n =3

θ = 36n + (-1)ⁿ6

θ = 36×3 - 6

θ = 102°

When n=4

θ = 36n + (-1)ⁿ6

θ = 36×4 + 6

θ = 150

When n=5

θ = 36n + (-1)ⁿ6

θ = 36×5 - 6

θ = 174°

When n = 6

θ = 36n+ (-1)ⁿ6

θ = 36×6 + 6

θ = 222°

When n = 7

θ = 36n + (-1)ⁿ6

θ = 36×7 - 6

θ = 246°

When n =8

θ = 36n + (-1)ⁿ6

θ = 36×8 + 6

θ = 294°

When n =9

θ = 36n + (-1)ⁿ6

θ = 36×9 - 6

θ = 318°

When n =10

θ = 36n + (-1)ⁿ6

θ = 36×10 + 6

θ = 366°

When n = 10 surpasses the θ range

Thus, the solutions range from n =0 to n=9

Therefore, there are 10 solutions within the interval 0<θ<2π

The question is missing some information. It should be phrased as follows:

<span><span>A container has 50 electronic components, with 10 identified as defective. If 6 components are randomly selected from the container, what is the probability that at least 4 of them are not defective? Additionally, if 8 components are drawn at random from the container, what is the probability that exactly 3 are defective?

</span>Answers
<span>Part 1.  0.02
Part 2. </span></span>0.0375<span><span>

</span>Explanation
Probability denotes the likelihood of an event occurring. It is computed as:
probability = (Number of favorable outcomes)/(Number of total outcomes)

Part 1
When 6 components are chosen, if 4 are confirmed functioning, then 2 must be defective.
P(at least 4 functional) = 4/40</span>× 2/10
                                            = 1/10 × 1/5
                                            = 1/50
                                            = 0.02

Part 2
Choosing 8 components, if 3 are defective, then 5 are functioning.
P(3 defective) = 3/40 × 5/10
                             = 15/400
                             = 3/80
                            = 0.0375

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.