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 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 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 |
In: Statistics and Probability
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 − 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. 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 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:
In: Computer Science
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), 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
OBSERVATIONS
| 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 |
| 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 |
| 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