The amount of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified at possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.2 ounces (these are the population parameters). Suppose a sample of 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags exceeded 10.6 ounces. (Hint: think of this in terms of a sampling distribution with sample size = 100)
In: Math
Two equal-length springs are “nested” together in order to form a shock absorber. If it is designed to arrest the motion of a 2-kg mass that is dropped above the top of the springs from an at-rest position, and the maximum compression of the springs is to be 0.2 m, determine the required stiffness of the inner spring, K B if the outer spring has a stiffness k A = 400 N/m
(s) is unknown, please will you show all the working out. Thank you
In: Mechanical Engineering
There are 2567 students enrolled at a small college, with 2053 of them enrolled in a sociology course.
In the sampling distribution of sample proportions of size 230, above what proportion will 52% of all sample proportions be?
Select all answers that apply to your calculation below. Use the z-table given below to answer the question:
| z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| -0.2 | 0.421 | 0.417 | 0.413 | 0.409 | 0.405 | 0.401 | 0.397 | 0.394 | 0.390 | 0.386 |
| -0.1 | 0.460 | 0.456 | 0.452 | 0.448 | 0.444 | 0.440 | 0.436 | 0.433 | 0.429 | 0.425 |
| -0.0 | 0.500 | 0.496 | 0.492 | 0.488 | 0.484 | 0.480 | 0.476 | 0.472 | 0.468 | 0.464 |
| z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 0.0 | 0.500 | 0.504 | 0.508 | 0.512 | 0.516 | 0.520 | 0.524 | 0.528 | 0.532 | 0.536 |
| 0.1 | 0.540 | 0.544 | 0.548 | 0.552 | 0.556 | 0.560 | 0.564 | 0.567 | 0.571 | 0.575 |
| 0.2 | 0.579 | 0.583 | 0.587 | 0.591 | 0.595 | 0.599 | 0.603 | 0.606 | 0.610 | 0.614 |
Select all that apply:
z=0.05
z=−0.01
p̂ =0.80
p̂ =0.32
p̂ =0.75
In: Statistics and Probability
You now set up a ligation reaction with the following components: • Plasmid Vector (0.05 µg/µL) • Insert DNA (0.2 µg/µL) • 10X Ligase Buffer • Water Because you are unsure what the best ratio of insert to plasmid is, as this is a new experiment for you, you plan to set up two experiments, one with a 3:1 ratio of insert to plasmid, and one with a 6:1 ratio of insert to plasmid. In both cases the amount of plasmid vector to be used in total will be 0.1 µg. The final reaction volume will be 20 µL. The final concentration of the 10X ligase buffer should be 1X. Using this information, complete the following chart regarding the volumes of each component used to prepare your samples. (Hint: For the plasmid and insert, determine the mass of DNA that needs to be in each tube using the information given, then use the equation C=mass/V to solve for the volume.) Ratio Volume of 0.05 µg/µL Plasmid Vector (µL) Volume of 0.2 µg/µL Insert DNA (µL) Volume of 10X Ligase Buffer (µL) Volume of Water (µL) Total Volume 3:1 20 µL 6:1 20 µL.
In: Biology
Suppose you and most other investors expect the inflation rate to be 7% next year, to fall to 5% during the following year, and then to remain at a rate of 3% thereafter. Assume that the real risk-free rate, r , will remain at 2% and that maturity risk premiums on Treasury securities rise from zero on very short-term securities (those that mature in a few days) to a level of 0.2 percentage points for 1-year securities. Furthermore, maturity risk premiums increase 0.2 percentage points for each year to maturity, up to a limit of 1.0 percentage point on 5-year or longer-term T-notes and T-bonds.
a. Calculate the interest rate on 1-, 2-, 3-, 4-, 5-, 10-, and 20-year Treasury securities, and plot the yield curve.
b. Now suppose ExxonMobil’s bonds, rated AAA, have the same maturities as the Treasury bonds. As an approximation, plot an ExxonMobil yield curve on the same graph with the Treasury bond yield curve. (Hint: Think about the default risk premium on ExxonMobil’s long-term versus short-term bonds.)
c. Now plot the approximate yield curve of Long Island Lighting Company, a risky nuclear utility.
In: Finance
Use the approximation that v→avg=p→f/m for each time step.
A paddle ball toy consists of a flat wooden paddle and a small
rubber ball that are attached to each other by an elastic band
(figure). You have a paddle ball toy for which the mass of the ball
is 0.014 kg, the stiffness of the elastic band is 0.890 N/m, and
the relaxed length of the elastic band is 0.325 m. You are holding
the paddle so the ball hangs suspended under it, when your cat
comes along and bats the ball around, setting it in motion. At a
particular instant the momentum of the ball is <−0.02, −0.01,
−0.02 > kg·m/s, and the moving ball is at location <−0.2,
−0.61, 0> m relative to an origin located at the point where the
elastic band is attached to the paddle.
(a) Determine the position of the ball 0.1 s later, using a Δt of 0.1 s. (Express your answer in vector form.)
(b) Starting with the same initial position (<−0.2, −0.61, 0> m) and momentum (<−0.02, −0.01, −0.02 > kg·m/s) determine the position of the ball 0.1 s later, using a Δt of 0.05 s. (Express your answer in vector form.)
In: Physics
The time series showing the sales of a particular product over
the past 12 months is contained in the Excel Online file below.
Construct a spreadsheet to answer the following
questions.
Use a=0.2 to compute the exponential smoothing forecasts for the time series (to 2 decimals).
Month |
Time-Series Value |
Forecast |
|---|---|---|
| 1 | 105 | |
| 2 | 130 | |
| 3 | 125 | |
| 4 | 100 | |
| 5 | 90 | |
| 6 | 120 | |
| 7 | 150 | |
| 8 | 135 | |
| 9 | 95 | |
| 10 | 75 | |
| 11 | 100 | |
| 12 | 105 | |
| 13 |
Use a smoothing constant of a=0.5 to compute the exponential smoothing forecasts (to 2 decimals).
Month |
Time-Series Value |
Forecast |
|---|---|---|
| 1 | 105 | |
| 2 | 130 | |
| 3 | 125 | |
| 4 | 100 | |
| 5 | 90 | |
| 6 | 120 | |
| 7 | 150 | |
| 8 | 135 | |
| 9 | 95 | |
| 10 | 75 | |
| 11 | 100 | |
| 12 | 105 | |
| 13 |
Compute MSE (to 2 decimals).
MSE ( a= 0.2 ) : (___)| Month | Time-Series Value |
| 1 | 105 |
| 2 | 130 |
| 3 | 125 |
| 4 | 100 |
| 5 | 90 |
| 6 | 120 |
| 7 | 150 |
| 8 | 135 |
| 9 | 95 |
| 10 | 75 |
| 11 | 100 |
| 12 | 105 |
| 13 |
In: Statistics and Probability
| The common stock of Escapist Films sells for $39 a share and offers the following payoffs next year: |
| Probability | Dividend | Stock Price | |
| Boom | 0.4 | $0 | $20 |
| Normal economy | 0.4 | $2 | $30 |
| Recession | 0.2 | $7 | $44 |
| Calculate the expected return and standard deviation of Escapist. (Round your answers to 2 decimal places. Use the minus sign for negative numbers if it is necessary.) |
| Expected rate of return % | |
| Standard deviation % |
| The common stock of Leaning Tower of Pita, Inc., a restaurant chain, will generate the following payoffs to investors next year: |
| Probability | Dividend | Stock Price | |
| Boom | 0.4 | $6 | $200 |
| Normal economy | 0.4 | $2 | $116 |
| Recession | 0.2 | $0 | $4 |
| The stock is selling today for $95. |
| Calculate the expected return and standard deviation of a portfolio half invested in Escapist and half in Leaning Tower of Pita. (Round your answers to 2 decimal places. Use the minus sign for negative numbers if it is necessary.) |
| Expected return of portfolio % | |
| Standard deviation of portfolio % |
| Why is the portfolio standard deviation lower than for either stock’s individually? | |
| The portfolio standard deviation is lower than for either stock’s individually because |
In: Finance
Parker Plastic, Inc., manufactures plastic mats to use with rolling office chairs. Its standard cost information for last year follows:
| Standard Quantity | Standard Price (Rate) | Standard Unit Cost | |||||||||||||||||||||||||||||||||||||||||||||||
| Direct materials (plastic) | 12 | sq ft. | $ | 0.63 | per sq. ft. | $ | 7.56 | ||||||||||||||||||||||||||||||||||||||||||
| Direct labor | 0.2 | hr. | $ | 10.10 | per hr. | 2.02 | |||||||||||||||||||||||||||||||||||||||||||
| Variable manufacturing overhead (based on direct labor hours) | 0.2 | hr. | $ | 0.60 | per hr. | 0.12 | |||||||||||||||||||||||||||||||||||||||||||
| Fixed manufacturing overhead $252,000 ÷ 840,000 units) | 0.30 | ||||||||||||||||||||||||||||||||||||||||||||||||
|
Parker Plastic
had the following actual results for the past year:
|
|||||||||||||||||||||||||||||||||||||||||||||||||
In: Accounting
Write a function ‘filter_wave(wave,n)’ and this will produce a new wave which is a smoothed version of the input ‘wave’ adhering to the following rules. The input parameter n will be explained later in part 2. • In the new wave, for every position i, it is the weighted sum of three positions of the original wave. More preciously, the new wave value in position i will be equal to new_wave[i] = wave[i-1] * 0.2 + wave[i]*0.6 + wave[i+1]*0.2 • Let len(wave) be L. The above calculation requires access to wave[-1] and wave[L] which do NOT exist in the original wave. You may assume that both wave[-1] and wave[L] are 0. • You should NOT modify the original wave input • And all the number in the new wave will be integers. You can simple use the function int() to convert any number into an integer. • Your function should return the new wave as a list.
Finally, modify the function ‘filter_wave(wave,n)’ for n ≥ 0. And this will repeat the filtering n times to the wave accumulatively and produce an even smoother wave. You still need to adhere to the rules mentioned in Part 1. Here is the expected wave for filter_wave(original_wave_sample,10)
In: Computer Science