We use partial fractions to find $A$ and $B$ such that:
\\[ \\simplify[std]{({a*a2+c*a1}*x+{c*b+a*d})/(({a1}x +{b})*({a2}x+{d}))} \\;\\;\\;=\\simplify[std]{ A/({a1}x+{b})+B/({a2}x+{d})}\\]
Multiplying both sides of the equation by $\\displaystyle \\simplify[std]{1/( ({a1}x+{b})({a2}x+{d}) )}\\;\\;$ we obtain:
\n$\\simplify[std]{A*({a2}x+{d})+B*({a1}x+{b}) = {a*a2+c*a1}*x+{a*d+c*b}} \\Rightarrow \\simplify[std]{({a2}A+{a1}B)*x+{d}*A+{b}*B={a*a2+c*a1}*x+{a*d+c*b}}$
\nIdentifying coefficients:
\nConstant term: $\\simplify[std]{ {d}*A+{b}*B={a*d+c*b} }$
\nCoefficent $x$: $ \\simplify[std]{ {a2}A+{a1}B = {a*a2+c*a1} }$
\nOn solving these equations we obtain $A = \\var{a}$ and $B=\\var{c}$
\nWhich gives:\\[ \\simplify[std]{({a*a2+c*a1}*x+{c*b+a*d})/(({a1}x +{b})*({a2}x+{d}))}\\;\\;= \\simplify[std]{{a}/({a1}x+{b})+{c}/({a2}x+{d})}\\]
\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\nSplit \\[\\simplify{({a*a2 + c*a1} * x + {a * d + c * b})/ (({a1}*x + {b}) * ({a2}*x + {d}))}\\] into partial fractions.
\nInput the partial fractions here: [[0]].
\n\n
\n ", "marks": 0, "gaps": [{"notallowed": {"message": "
Input as the sum of partial fractions.
", "showStrings": false, "strings": [")(", ")*("], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": [], "checkingaccuracy": 1e-05, "vsetrange": [10, 11], "showpreview": true, "marks": 2, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a} / ({a1}*x + {b}) + ({c} / ({a2}*x + {d}))", "checkingtype": "absdiff", "checkvariablenames": false, "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "\n\n
\n ", "variable_groups": [], "progress": "ready", "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "random(-9..9 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "random(-9..9 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "random(-9..9 except [0,round(b*a2/a1)])", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "nb": {"definition": "if(c<0,'taking away','adding')", "templateType": "anything", "group": "Ungrouped variables", "name": "nb", "description": ""}, "a1": {"definition": "random(1..8)", "templateType": "anything", "group": "Ungrouped variables", "name": "a1", "description": ""}, "a2": {"definition": "random(1..8)", "templateType": "anything", "group": "Ungrouped variables", "name": "a2", "description": ""}, "s1": {"definition": "if(c<0,-1,1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s1", "description": ""}}, "metadata": {"notes": "\n \t\t \t\t
5/08/2012:
\n \t\t \t\tAdded tags.
\n \t\t \t\tAdded description.
\n \t\t \t\tChanged to two questions, for the numerator and denomimator, rather than one as difficult to trap student input for this example. Still some ambiguity however.
\n \t\t \t\t12/08/2012:
\n \t\t \t\tBack to one input of a fraction and trapped input in Forbidden Strings.
\n \t\t \t\tUsed the except feature of ranges to get non-degenerate examples.
\n \t\t \t\tChecked calculation.OK.
\n \t\t \t\tImproved display in content areas.
\n \t\t \n \t\t", "description": "Split $\\displaystyle \\frac{ax+b}{(cx + d)(px+q)}$ into partial fractions.
"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "cormac's copy of BS2.1", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "cormac murphy", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/26/"}], "tags": ["Probability", "probability", "statistics", "udf"], "progress": "testing", "metadata": {"notes": "\n\t\t \t\t28/12/2012:
\n\t\t \t\tUsing the inbuilt table function for now. This needs to be changed - either to direct input of an html table or improving the table function e.g. adding borders etc.
\n\t\t \t\tThe udf accumdisp(a,t) outputs a string of the form a[0]+a[1]+..a[t-1] - useful to show in the solution the elements of the list we are summing over.
\n\t\t \t\tThere is a scenario variable sk, which is intended to be the beginning of a list of possible randomised scenarios. Probably best if this included other text based string variables (e.g. car loans could be the value of such a variable).
\n\t\t \t\tEasy to make this have a variable number of ranges of loans. Only need to pay some attention to the creation of the list n giving the number of loans in each range - need to make that sensible.
\n\t\t \n\t\t", "description": "Simple probability question. Counting number of occurrences of an event in a sample space with given size and finding the probability of the event.
"}, "statement": "\n\t{sc[k]}
\n\t{table(data,[' From',' To', ' Loans Made'])}
\n\t\n\t \n\t \n\t", "advice": "\n\t
a) The number of loans less than £$\\var{u1}$ is $\\var{accumdisp(n,t)}=\\var{sum(n[0..t+1])}$
\n\tSince there are $\\var{thismany}$ loans the probability of choosing one of these loans is $\\displaystyle \\frac{\\var{sum(n[0..t+1])}}{\\var{thismany}}=\\var{ans1}$ to 2 decimal places.
\n\tb) The number of loans greater than £$\\var{o1}$ is $\\var{accumdisp(n[v+1..abs(n)],abs(n)-v-2)}=\\var{sum(n[v+1..abs(n)])}$.
\n\tSince there are $\\var{thismany}$ loans the probability of choosing one of these loans is $\\displaystyle \\frac{\\var{sum(n[v+1..abs(n)])}}{\\var{thismany}}=\\var{ans2}$ to 2 decimal places.
\n\tc) There are $\\var{accumdisp(n[p+1..q+1],q-p-1)}=\\var{sum(n[p+1..q+1])}$ loans between £$\\var{a[p]}$ and £$\\var{a[q]-1}$.
\n\tHence the probability of selecting one of these loans in this range for review is $\\displaystyle \\frac{\\var{sum(n[p+1..q+1])}}{\\var{thismany}}=\\var{ans3}$ to 2 decimal places.
\n\t \n\t \n\t", "rulesets": {}, "variables": {"k": {"name": "k", "definition": "random(0..abs(sc)-1)"}, "sc": {"name": "sc", "definition": "['A bank made '+{thismany}+' car loans last year. The amounts were as follows (?):']"}, "thismany": {"name": "thismany", "definition": "random(300..1000#100)"}, "data": {"name": "data", "definition": "\n\t\t [[0,a[0]-1,n[0]],\n\t\t [a[0],a[1]-1,n[1]],\n\t\t [a[1],a[2]-1,n[2]],\n\t\t [a[2],'plus',n[3]]]\n\t\t \n\t\t \n\t\t \n\t\t"}, "a": {"name": "a", "definition": "[a0,a0+b0,a0+2*b0]"}, "a0": {"name": "a0", "definition": "random(1000..4000#1000)"}, "b0": {"name": "b0", "definition": "random(1000..3000#1000)"}, "n": {"name": "n", "definition": "[n0,n1,thismany-n0-n1-n3,n3]"}, "n0": {"name": "n0", "definition": "round(thismany/random(15,25))"}, "n1": {"name": "n1", "definition": "round(thismany/random(3,6))"}, "n3": {"name": "n3", "definition": "round(thismany/random(11,14))"}, "u1": {"name": "u1", "definition": "a[t]"}, "ans1": {"name": "ans1", "definition": "precround(sum(n[0..t+1])/thismany,2)"}, "o1": {"name": "o1", "definition": "a[v]"}, "ans2": {"name": "ans2", "definition": "precround((thismany-sum(n[0..v+1]))/thismany,2)"}, "ans3": {"name": "ans3", "definition": "precround((n[1]+n[2])/thismany,2)"}, "t": {"name": "t", "definition": "random(0..abs(a)-1)"}, "v": {"name": "v", "definition": "random(0..abs(a)-1 except t)"}, "p": {"name": "p", "definition": 0.0}, "q": {"name": "q", "definition": 2.0}}, "functions": {"accumdisp": {"parameters": [["a", "list"], ["k", "number"]], "type": "string", "language": "jme", "definition": "if(k=0,'$\\\\var{a[0]}$','$\\\\var{a[0]}$ + '+accumdisp(a[1..abs(a)],k-1))"}}, "parts": [{"type": "gapfill", "marks": 0.0, "prompt": "\n\t\t\tOne of these loans is sampled randomly for review by the bank. What is the probability that it is :
\n\t\t\ta) Under £$\\var{u1}$? Probability = ? [[0]] (answer to 2 decimal places).
\n\t\t\tb) Over £$\\var{o1-1}$? Probability = ? [[1]] (answer to 2 decimal places).
\n\t\t\tc) Between £$\\var{a[p]}$ and £$\\var{a[q]-1}$? Probability = ? [[2]] (answer to 2 decimal places).
\n\t\t\t\n\t\t\t
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dfg
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