Questions
1.An airoplane with four engines is able to fly if at least one engine is functioning...

1.An airoplane with four engines is able to fly if at least one engine is functioning on each wing. Let Ai be the event that the ith engine is working, i = 1, 2, 3, 4, where engines 1 and 2 are on the left wing and engines 3 and 4 are on the right wing. Let A be the event that the airoplane is able to fly. Express A in terms of A1, . . . , A4, using complements, intersections and/or unions.

2. Consider the airplane from question 1. Assume that all engines have a probability of 0.1 of failing, and that they fail independently of each other. What is the probability that the plane will be able to fly?

In: Statistics and Probability

Question Regarding RFLP marker I have an RFLP marker for a family with 2 parents and...

Question Regarding RFLP marker

I have an RFLP marker for a family with 2 parents and 4 children. The gender of the parents are not told

1. Child 1 has 3 bands, how to determine if that child has trisomy of the chromosome due to a non-disjunction during Meiosis II or Meiosis I and how to tell which parents gametes caused this parent 1 or parent 2.

2. How to determine which of the children is a half sibling with only 1 biological parent

3. Child # 4 is missing information, how to tell which information that child is missing

In: Biology

Using repeated measure on SPSS - I need tests of within and pairwise comparisons ( Upload...

Using repeated measure on SPSS - I need tests of within and pairwise comparisons

( Upload screenshot of SPSS - tests of within and pairwise comparisons )

In an effort to understand test anxiety better, an educational researcher attempts to operationalize anxiety by informing subjects that completing a task in the time limit is required if the subject is to receive a reward. Each subject receives four sets of 10 simple calculation problems. Each set imposes a more stringent time requirement than the one preceding after which anxiety is gauged. Anxiety scores for 8 subjects are as follows:

Time 1

Time 2

Time 3

Time 4

1

8

8

9

9

2

7

7

8

10

3

4

4

4

5

4

2

3

5

5

5

5

6

6

8

6

5

5

7

9

7

4

4

5

4

8

2

3

6

7

  1. Does anxiety appear to be related to time limits?
  2. Where are the significant differences?

In: Statistics and Probability

You are given two integer arrays a and b of the same length. Let's define the...

You are given two integer arrays a and b of the same length. Let's define the difference between a and b as the sum of absolute differences of corresponding elements:

difference = |a[0] - b[0]| + |a[1] - b[1]| + ... + |a[a.length - 1] - b[b.length - 1]|

You can replace one element of a with any other element of a. Your task is to return the minimum possible difference between a and b that can be achieved by performing at most one such replacement on a. You can also choose to leave the array intact.

Example:

For a = [1, 3, 5] and b = [5, 3, 1], the output should be minDiffOfArrays(a, b) = 4.
If we leave the array a intact, the difference is |1 - 5| + |3 - 3| + |5 - 1| = 8;
If we replace a[0] with a[1], we get a = [3, 3, 5] and the difference is |3 - 5| + |3 - 3| + |5 - 1| = 6;
If we replace a[0] with a[2], we get a = [5, 3, 5] and the difference is |5 - 5| + |3 - 3| + |5 - 1| = 4;
If we replace a[1] with a[0], we get a = [1, 1, 5] and the difference is |1 - 5| + |1 - 3| + |5 - 1| = 10;
If we replace a[1] with a[2], we get a = [1, 5, 5] and the difference is |1 - 5| + |5 - 3| + |5 - 1| = 10;
If we replace a[2] with a[0], we get a = [1, 3, 1] and the difference is |1 - 5| + |3 - 3| + |1 - 1| = 4;
If we replace a[2] with a[1], we get a = [1, 3, 3] and the difference is |1 - 5| + |3 - 3| + |3 - 1| = 6;
The final answer is 4, since it's the minimum possible difference.

In: Computer Science

1) The position of an object moving along a straight line is s(t) = t^3 −...

1) The position of an object moving along a straight line is s(t) = t^3 − 15t^2 + 72t feet after t seconds. Find the object's velocity and acceleration after 9 seconds.

2) Given the function f (x ) =−3 x 2 + x − 8 ,
(a) Find the equation of the line tangent to f(x ) at the point (2, −2) .

(b) Find the equation of the line normal to f(x ) at the point (2, −2)

3) Differentiate the following functions.
(a) f (t) = 5t^3 lnt
(b) y =x^2/2 x + 1

4) Differentiate the following function.
f (x ) =sin x/1 + cos x

5) Differentiate the following functions.
(a) g ( x) = ( 4 x^2 + 3)^8
(b) f (x ) = cos( 7e^x − 4)

6) Differentiate y =cos5 x/x^2

7) Differentiate the following functions.
(a) g ( x) = ln( 5 x^2 −3)
(b) y = ln(cos x)

In: Math

Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the given sample data....

Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the given sample data. An experiment was conducted to determine whether a deficiency of carbon dioxide in the soil affects the phenotype of peas. Listed below are the phenotype codes where 1 equals smooth dash yellow​, 2 equals smooth dash green​, 3 equals wrinkled dash yellow​, and 4 equals wrinkled dash green. Do the results make​ sense? 2 2 1 4 3 1 4 1 4 2 1 1 2 1 ​(a) The mean phenotype code is nothing. ​(Round to the nearest tenth as​ needed.) ​(b) The median phenotype code is nothing. ​(Type an integer or a​ decimal.) ​(c) Select the correct choice below and fill in any answer boxes within your choice. A. The mode phenotype code is nothing. ​(Use a comma to separate answers as​ needed.) B. There is no mode. ​(d) The midrange of the phenotype codes is nothing. ​(Type an integer or a​ decimal.) Do the measures of center make​ sense? A. All the

measures of center make sense since the data is numerical. B. Only the mode makes sense since the data is nominal. C. Only the​ mean, median, and midrange make sense since the data is nominal. D. Only the​ mean, median, and mode make sense since the data is numerical.

In: Statistics and Probability

Create a Packet Tracer file that completes the required steps below. Add two Cisco 2960 switches...

Create a Packet Tracer file that completes the required steps below.

Add two Cisco 2960 switches and wire it up accordingly; use a crossover cable.

Device

Interface

To

Switch 1

G0/0

Switch 2

G0/1

Switch 2

G0/1

Switch 1

G0/0

After you have added the switches, add the following devices and wire accordingly.

Device

Interface

To

PC1-A

F0

Switch 1

F0/4

PC1-B

F0

Switch 1

F0/3

PC2-A

F0

Switch 2

F0/4

PC2-B

F0

Switch 2

F0/3

Configure the following information on the computers:

Device

IP

Subnet

PC1-A

10.10.0.145

255.255.255.0

PC1-B

10.10.0.133

255.255.255.0

PC2-A

10.10.0.243

255.255.255.0

PC2-B

10.10.0.210

255.255.255.0

Configure the following items:

  • Set passwords for the Console and VTY 0 4 lines.
  • The password should be set for login and the password should be "P@ssword#1".
  • Set a enable password that is a secret; the password should be "P@ssword#2".
  • Set a banner that is appropriate.
  • Set the hostname as appropriate.
  • Enable the service for securing passwords.

In: Computer Science

Explanation of the problem chapter 3.12 problem 49 of the book introduction to the mathematical programming...

Explanation of the problem chapter 3.12 problem 49 of the book introduction to the mathematical programming 4th edition
Solution: Step # 1
Let Xij be the amount of money invested at the beginning of month i, for a period of j month.
Objective function: Step # 2
The objective is to maximize the available cash at the beginning of month 5.
X14 = collect the money invested at the beginning of month 1 of 4 months.
X23 = collect the money invested at the beginning of month 2 of 3 months.
X32 = collect the money invested at the beginning of month 3 of 2 months.
X41 = collect the money invested at the beginning of month 4 of 1 month.
Therefore the objective function is:
Maximize Z = 1.08 X14 + 1.03 X23 + 1.01 X32 + 1.001 X41
Restriction 1
It would be the money invested at the beginning of month 1 plus bills paid in month 1 which would be equal to the money available at month 1.

                       X11 + X12 + X13 + X14 + 600 = 400 + 400 (MONTH 1)
Restriction 2
It would be the money invested at the beginning of month 2 plus bills paid in month 2 which would be equal to the money available at month 2.

X21 + X22 + X23 + 500 = 1.001 X11 + 800 (MONTH 2)
Restriction 3 and 4
In the same way for the rest of the month we have:

X31 + X32 + 500 = 1.001 X12 + 1.001 X21 + 300 (MONTH 3)
X41 + 250 = 1.001 X13 + 1.001 X22 + 1.001 X31 + 300 (MONTH 4)

What I need is an explanation of the problem of how the data was extracted at each step

In: Advanced Math

The following table lists the components needed to assemble an end item, lead times (in weeks),...

The following table lists the components needed to assemble an end item, lead times (in weeks), and quantities on hand.

Item Lead Time Amount
on Hand
Direct Components
End 3 0 L(2), C(1), K(3)
L 3 12 B(2), J(3)
C 4 16 G(2), B(2)
K 4 21 H(4), B(2)
B 3 26
J 4 33
G 4 4
H 3 0


a. If 43 units of the end item are to be assembled, how many additional units of B are needed? (Hint: You don’t need to develop an MRP plan.)

Additional units            

In: Operations Management

2. Based off your data, what are the standard half-cell potentials for Silver and Iron/Platinum? Explain your reasoning.

 

OBSERVATIONS

Part 1 : Electrochemical Cell Voltages (0.1 M & 0.2 M Solutions)
Trial Right Cell Metal Right Cell Solution Left Cell Metal Left Cell Solution Total Voltage
#1 Lead 4 mL of 0.1 M Lead Nitrate [Pb(NO3)2] Copper 40 mL of 0.1M Copper Nitrate [Cu(NO3)2] 0.47 v
#2 Lead 4 mL of 0.1M Lead Nitrate [Pb(NO3)2] Silver 40 mL of 0.1M Silver Nitrate [AgNO3] 0.90 v
#3 Lead 4 mL of 0.1MLead Nitrate [Pb(NO3)2] Platinum 20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4] 0.91 v
#4 Copper 4 mL of 0.1M Copper Nitrate [Cu(NO3)2] Silver 40 mL of 0.1M Silver Nitrate [AgNO3] 0.43 v
#5 Copper 4 mL of 0.1M Copper Nitrate [Cu(NO3)2] Platinum 20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4] 0.44 v
#6 Silver 4 mL of 0.1M Silver Nitrate [AgNO3] Platinum 20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4] -0.02 v
Part 2 : Electrochemical Cell Voltages : (1.0M Solutions)
Trial Right Cell Metal Right Cell Solution Left Cell Metal Left Cell Solution Total Voltage
#1 Lead 4 mL of 1.0M Lead Nitrate [Pb(NO3)2] Copper 40 mL of 1.0M Copper Nitrate [Cu(NO3)2] 0.47 v
#2 Lead 4 mL of 1.0M Lead Nitrate [Pb(NO3)2] Silver 40 mL of 1.0M Silver Nitrate [AgNO3] 0.93 v
#3 Lead 4 mL of 1.0M Lead Nitrate [Pb(NO3)2] Platinum 20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4] 0.91 v
#4 Copper 4 mL of 1.0M Copper Nitrate [Cu(NO3)2] Silver 40 mL of 1.0M Silver Nitrate [AgNO3] 0.46 v
#5 Copper 4 mL of 1.0M Copper Nitrate [Cu(NO3)2​] Platinum 20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4​] 0.44 v
#6 Silver 4 mL of 1.0M Silver Nitrate [AgNO3] Platinum 20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4​] -0.02 v
Part 3 : Variation of Concentration
Trial Right Cell Metal Right Cell Solution Left Cell Metal Left Cell Solution Total Voltage
#1 Silver 4 mL of 0.1M Silver Nitrate [AgNO3] Silver 40 mL of 1.0M Silver Nitrate [AgNO3] 0.06 v
#2 Silver 4 mL of 0.1M Silver Nitrate [AgNO3​] Silver 40 mL of 0.1M Silver Nitrate [AgNO3] 0.00 v
#3 Silver 4 mL of 0.1M Silver Nitrate [AgNO3​] Silver 40 mL of 0.1M Silver Nitrate [AgNO3] & 400 mL Water -0.06 v

CONCLUSIONS:

1. When comparing the Part 1 and Part 2 data, how did the voltages differ when varying concentrations?

4 of the trials voltage charge remained the same, while 2 of the trials went up in voltage, increased by 0.03 v.

2. Based off your data, what are the standard half-cell potentials for Silver and Iron/Platinum? Explain your reasoning.

3. When comparing your data in Part 3, how did varying the concentration affect the voltage? Which trial had the largest voltage? Which trial had the lowest voltage?

The voltage decreased with each variation. Trial #1 had the largest voltage and trial #3 had the lowest voltage.

4. In the following reactions, determine which element is oxidized and which one is reduced.

- Sn + 4 HNO3 ------- SnO2 + 4 NO2 + 2 H2O

Oxidized Element: Sn ------- SnO2

Reduced Element: 4HNO3 ------ 4NO2

- Mg + Br2 ---------- MgBr2

Oxidized Element: Mg ----- Mg

Reduced Element: 2Br ------ 2Br

In: Chemistry