University:
University of Michigan
Course:
MATH 115 | Calculus I
Academic year:

2024
Views:

462

Pages:

3
Author:

Luciano Kirby

Question 1 Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a particular type, its mean value and standard deviation are 70 GPa and 1.6 GPa, respectively (values given in the article "Influence of Material Properties Variability on Springback and Thinning in Sheet Stamping Processes: A Stochastic Analysis" (Intl. J. of Advanced Manuf. Tech., 2010: 117-134)). If X is the sample mean Young's modulus for a random sample of n = 16 sheets, where is the sampling distribution of X centered, and what is the standard deviation of the X distribution? Answer μ΄α = 70GPa qz = 0.4GPa Question 2 What is the future of 2200 i n v e s t e d f o r n i n e m o n t h s a t 4.25 S e l e c t o n e : α : 2200investedforninemonthsat4.25Selectone:α: 2270 b: 2341.73 c : 2341.73c: 2234>97 d: 2269.76 A n s w e r c o m p o u n d i n t e r e s t f o r m u l a p = a ( 1 + r / n ) n t W h e r e A = p r e s e n t v a l u e o f i n v e s t m e n t P = f u t u r e v a l u e r = r a t e o f i n t e r e s t n = n u m b e r o f t i m e s i n t e r e s t c o m p o u n d i n a p e r i o d t = t i m e p e r i o d G i v e n A = 2269.76Answercompoundinterestformulap=a(1+r/n)ntWhereA=presentvalueofinvestmentP=futurevaluer=rateofinterestn=numberoftimesinterestcompoundinaperiodt=timeperiodGivenA= 2200 r=4.25%=0.0425 n=2 of compound semi annually t = (9/12)yrs p = 2200(1 + (0.0425/2)(29/(12)) p= 2270.5 Q u e s t i o n 3 A p a y m e n t o f 1500 2270.5Question3Apaymentof1500 is due in three months.Find the equivalent value at nine months if the interest rate is 4% select one α: 1500 b : 1500b: 1530 c: 1560 d : 1560d: 1470 Answer Given interest rate is 4% compounded quarterly F.V = P.V(1+r/k)nk F. V = Future value P. V = Present value r =rate of interest k =number of times compounded per year n =number of years steps: r=4 k = 4(since compounded quarterly) n = 3 months = 3/12 years = (1/4)years nk = 4 * 1/4 = 1 substitute values in the above equation P. V = 1500(1+ (0.04/4)) = 1485.5 Find the equivalent value r=4 k = 4(since compounded quarterly) n = 9 months= 9/12 P.V = $1485.15 F.V = P. V(1+(r/k)nk) F. V = 1485.5(1+(0.04/4))3/4 * 4 = $1530.15 Option B is correct Question 4 The solid lies between the Planes perpendicular to the x axis at X=-lamd X=1. Cross- sections perpendicular to the x axis between between these planes are circular disk with the diameters run from the semicircle y =-root(1-x2)to the semi circle y=root(1-x*2).Find a formula for the area of cross section A(x) Answer The bounds of the region are x -1 and x =1 so integrate the area function from -1 to 1 to find the volume. V = ∫₁₋₁ A(x)dx V = ∫₁₋₁ 4(1 - x²)dx = 4(x- x³/3)|¹₋₁ = 4(1 - (1)³/3) - 4(−1 - (-1)³/3) = 4(1-1/3) - 4(-1+1/3) = 4(2/3)-4(-2/3) = 8/3 + 8/3 = 16/3