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Message: <blockquote data-quote="NRTG" data-source="post: 28925539" data-attributes="member: 573158">1.  The relation R is defined on Z by aRb if a+b is even. To show that R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive.Reflexive: For all a in Z, a+a=2a is even. Therefore, aRa for all a in Z.Symmetric: For all a,b in Z, if aRb then bRa. If a+b is even then b+a is also even. Therefore, if aRb then bRa.Transitive: For all a,b,c in Z, if aRb and bRc then aRc. If a+b is even and b+c is even then (a+b)+(b+c) = (a+c)+2b is even. Therefore, if aRb and bRc then aRc.2. (a)The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.In this case, we have P = $6000, A = $16000, n = 1 (compounded annually), and t = 10. We need to find r.Substituting these values into the formula and solving for r, we get:16000 = 6000(1 + r/1)^(1*10) 16000/6000 = (1 + r)^10 2.6667 = (1 + r)^10 log(2.6667) = log(1 + r)^10 log(2.6667) = 10log(1 + r) log(2.6667)/10 = log(1 + r) 0.0083 = log(1 + r) 10^0.0083 - 1 = rTherefore the annual interest rate is approx 8.3 %2.(b)We can use the formula for compound interest to solve this problem. The formula is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.In this case, we have P = $6000, A = $40000, n = 1 (compounded annually), and r = 8.3% (from previous question). We need to find t.Substituting these values into the formula and solving for t, we get:40000 = 6000(1 + 0.083/1)^(1t) 40000/6000 = (1 + 0.083)^t 6.6667 = (1.083)^t log(6.6667) = log(1.083)^t log(6.6667) = tlog(1.083) log(6.6667)/log(1.083) = tTherefore, it will take approximately 21 years for the investment to be worth at least $40000 .</blockquote>

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