Calculate the missing amounts for each of the following

Calculate the missing amounts for each of the following firms: (Do not round intermediate calculations and round selling price answers to 2 decimal places.)

Solution

Particulars Firm A Firm B Firm C Firm D Units Sold (Total cont/cont per unit)                    11,200.00         8,400.00                 3,500.00                 4,720.00 Selling Price per unit(VC+Cont)                            24.00               29.80                         7.30                       59.95 VC per unit (SP-cont)                            15.00               18.20                         4.20                       51.25 Cont Per Unit(SP-VC)or Total cont███████████ ███████                               ███████████               ███████████                         ████████                          ███████████ ████████ ███████████(█████+████ ████████)                  ████,██████████       ██,████████████              ████,██████████████               ███,███████████████ ████ = █████ ███████ &#████████; ██████████                    ██,█████████████       █████,████████████              ████,███████████               █████,██████████████ ███████████████████ ████ (███████████████)                    █████,█████████       ███,███████              (█,█████████████)               (██,██████████)

. A market research firm is investigating the appeal of

. A market research firm is investigating the appeal of three package designs. The table below gives information obtained through a sample of 200 consumers. The three package designs are labeled A, B, and C. The consumers are classified according to age and package design preference. A B C Total Under 25 years 22 34 40 96 25 or older 54 28 22 104 Total 76 62 62 200 If one of these consumers is randomly selected, what is the probability that the person prefers design A? (Points : 4) 0.76 0.38 0.33 0.22 0.39 2. Let A be the event that a student is enrolled in an accounting course, and let S be the event that a student is enrolled in a statistics course. It is known that 30% of all students are enrolled in an accounting course and 40% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and accounting. Find the probability that a student is in accounting and is also in statistics. (Points : 4) 0.15 0.70 0.55 0.12 0.60 3. If X and Y are mutually exclusive events, then if X occurs _______. (Points : 4) Y must also occur Y cannot occur X and Y are independent X and Y are complements A and Y are collectively exhaustive 4. The area to the left of the mean in any normal distribution is equal to _______. (Points : 4) the mean 1 the variance 0.5 -0.5 5. If x is a normal random variable with mean 80 and standard deviation 5, the z-score for x = 88 is ________. (Points : 4) 1.8 -1.8 1.6 -1.6 8.0 6. Within a range of z scores from -1 to +1, you can expect to find _______ per cent of the values in a normal distribution. (Points : 4) 95 99 68 34 100 7. The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. What is the probability that a randomly selected bulb would last fewer than 1100 hours? (Points : 4) 0.4772 0.9772 0.0228 0.5228 0.5513 8. Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. To be 99% sure that we will not be late in completing the project, we should request a completion time of _______ work-days. (Points : 4) 211 207 223 200 250 9. Financial analyst Larry Potts needs a sample of 100 securities listed on either the New York Stock Exchange (NYSE) or the American Stock Exchange (AMEX). According to the Wall Street Journal\’s \”Stock Market Data Bank,\” 2,531 NYSE securities and AMEX 746 securities were traded on the previous business day. Larry directs his staff to randomly select 77 NYSE and 23 AMEX securities. His sample is a ____________. (Points : 4) disproportionate systematic sample disproportionate stratified sample proportionate stratified sample proportionate systematic sample proportionate cluster sampling 10. Suppose 40% of a student population possess works part time. If a random sample of size 300 is drawn from the population, then the probability that 44% or fewer of the samples works part time is _______. (Points : 4) 0.0793 0.4207 0.9207 0.9900 1.0000

Solution

it is chegg rule to ask one question in one post. So, i am answering only one quaestion.

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Find the polynomial. Degree 5 roots -2, 1(multiplicity

Find the polynomial. Degree 5 roots: -2, 1(multiplicity 2), 4i

Please show all steps. Thanks

Solution

Degree 5 roots: -2, 1(multiplicity 2), 4i

roots are -2 , 1 ,1,4i. it 4i .If 4i is on root then its complex conjugate should also ███ ██████ █████████ ███████ █████

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The hint of the question is helpful in the sense that I

The hint of the question is helpful in the sense that Inow know that E(N)= e = taylor series. But I am not sure how to usethe hint to answer this question. Does anyone know how to do thisquestion?

Solution

Use E(N) = e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ……&#█████;██████

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Assignment 8 Question 20 Give an equation of the form f

Assignment 8

Question 20

Give an equation of the form f (x) = Asin(B xC) + D which could be used to represent the given graph. (Note: C or D may be zero.)

Solution

f(x) = Asin(B x – C) + D

A = Amplitude : from the graph Amplitude = (3+1)/2

==> A = 2

function resembles sine function with a upward vertical shift of 1 unit

==> D = 1

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An athletic director found that 31 of the football play

An athletic director found that 31% of the football players have an A average in school. If 2% of the students at the school play football, what is the probability that a student chosen at random will be a football player with an A average?

Please show work

Solution

Consider there are 5000 students in total

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Maria and Zoe are taking Biology 105 but are in differe

Maria and Zoe are taking Biology 105 but are in different classes. Maria\’s class has an average of 78% with a standard deviation of 5% on the midterm, whereas Zoe\’s class has an average of 83% with a standard deviation of 12%. Assume that scores in both classes follow a normal distribution.

Convert Maria\’s midterm score of 84 to a standard Z score.

Convert Zoe\’s midterm score of 89 to a standard Z score.

Who did better relative to her class?

Solution

maria z=84-78/5=1.2

zoe z=89-83/12=0.5

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