intermediate algebra and pre statistic test in morning 8am
Math 30+ Review #4 Solutions for selected exercises Instructions for the exam Show all work for full credit. Clearly circle or box your answers when applicable. When rounding final answers, use three decimal places (unless there are fewer than 3, or unless 3 decimals is inappropriate for the context of the problem). You may use a pencil, eraser, ruler (or other straight edge), scientific calculator, StatCrunch, the e-text, and desmos.com. PS 8.1, 8.2, and 8.3 3) Solve ??−? ? = ??+? ? Lcd is 6: ? ( ??−? ? ) = ? ( ??+? ? ) ?(?? − ?) = ?(?? + ?) ?? − ? = ?? + ? ?? − ?? = ? + ? ? = ? Check: ?(?)−? ? = ?(?)+? ? ? … 5) The sales of musical instruments were $?. ?? million in 2010, and they decreased by about $?. ?? million per year until 2014. a) Find an equation of a linear model that describes the situation. Explain what the variables represent. (Hint: one of the variables will have to represent the number of years since 2010). Let ? = number of years since 2010, and let ?(?) = sales ? years after 2010, then ?(?) = ?? + ? ? = 0 year 2010, when sales = $6.39 million y-intercept = (0, 6.39) b = 6.39 Decrease of $0.28 million r.o.c. = m = -0.28 Equation: ?(?) = −?. ??? + ?. ??, in millions of dollars b) Predict the amount of sales in 2017. 2017 ? = ? ?(?) = −?. ??(?) + ?. ?? = ?. ?? sales were $4.43 million in 2017 c) Estimate in which year sales were $?. ? million. ?. ? = −?. ??? + ?. ?? −?. ?? = −?. ??? ?. ?? … = ? So ? ≈ ?. ? ? ≈ ?. ? years since 2010 puts it right at the end of 2013, but this still means it hits the sales of $5.3 million in the year 2013. IA 1.3 6) Suppose that the percentages of men and women, ?(?) and ?(?) respectively, who have completed for or more years of college are modeled by the system ? = ?(?) = ?. ??? + ??. ?? ? = ?(?) = ?. ??? + ??. ?? where ? is the number of years since 1990. a) Use substitution or elimination to estimate when the percentage of women who have completed four or more years of college was equal to the percentage of men who had completed four or more years of college. Ie: find ? such that the expressions ? = ?(?) and ? = ?(?) are equal. To do this, substitute the expression for ?(?) in for ? in the equation ? = ?(?): ?(?) = ?(?) Then substitute in the respective equations: ?. ??? + ??. ?? = ?. ??? + ??. ?? And solve for ?: ?. ??? − ?. ??? = ??. ?? − ??. ?? ?. ??? = ?. ?? ? = ??. ?? … ≈ ??. ? About 22.9 years after 1990 is towards the end of 2012. So the percentage of women who completed least 4 years of college equaled the percentage of men who completed at least 4 years of college somewhere near the end of 2012. b) What was that percentage? ?(??. ?) = ?(??. ?) = ?. ??(??. ?) + ??. ?? = ??. ? So about 31.4% of both men and women had completed at least 4 years of college by the end of 2012. PS 8.4 & IA 3.5 (IA 3.5 was absorbed into PS 8.4 lecture) 7) Solve the following formulas for the specified variable. a) ?(? ?? ?) = ?(?) + ?(?) − ?(? ??? ?); for ?(?) ?(? ?? ?) − ?(?) + ?(? ??? ?) = ?(?) + ?(?) − ?(?) − ?(? ??? ?) + ?(? ??? ?) ?(? ?? ?) − ?(?) + ?(? ??? ?) = ?(?) b) ?? = ?? √?−? ; for ? ?? = ?? √?−? (√? − ?) ∙ ?? = (√? − ?) ∙ ?? √?−? (√? − ?) ∙ ?? = ?? √? − ? = ?? ?? (√? − ?) ? = ( ?? ?? ) ? ? − ? = ( ?? ?? ) ? ? = ( ?? ?? ) ? + ? PS 8.5 9) Solve the compound inequality ? ≤ ?(? − ?) ≤ ?. Describe the solution set as an inequality, in interval notation, and on a graph. ? ≤ ?(? − ?) ≤ ? ? ≤ ? − ?? ≤ ? −? ≤ −?? ≤ −? −? −? ≥ ? ≥ −? −? Inequality: ? ? ≥ ? ≥ ? ? or, equaivalently, ? ? ≤ ? ≤ ? ? Interval: [ ? ? , ? ? ] Graph: IA 4.1 10) Solve ? − |??? + ? ? | = −? ? − |??? + ? ? | = −? − |??? + ? ? | = −? |??? + ? ? | = ? …so ??? + ? ? , whatever it is, is a distance of 6 away from 0 on the number line. Only two numbers are 6 away from 0 on the number line, namely -6 and 6. So… ??? + ? ? = −? ??? + ? = −?? ??? = −?? ? = − ?? ?? Or Or ??? + ? ? = ? ??? + ? = ?? ??? = ? ? = ? ?? = ? ? Check by plugging both solutions back in to the original equation or by graphing. 13) Solve |?? + ?| − ? < .="" describe="" the="" solution="" set="" as="" an="" inequality,="" in="" interval="" notation,="" and="" on="" a="" graph.="" |??="" +="" |="" −="">< |??="" +="" |="">< whatever="" +="" is,="" its="" distance="" from="" 0="" along="" the="" number="" line="" is="" less="" than="" 12,="" so="" +="" can="" be="" anything="" between="" -12="" and="" 12,="" exclusive.="" so…="" −??="">< +="">< −??=""><>< −=""><>< inequality:="" −=""><>< ? interval: (− ?? ? , ?) graph: ia 1.6 (no systems) 17) graph the linear inequality in two variables by hand: ?(? − ?) + ? ≤ −? solve the inequality for ?: ?(? − ?) + ? ≤ −? ?? − ? + ? ≤ −? ? ≤ −?? + ? so the equation for the boundary line is ? = −?? + ? and, since the inequality ≤ allows for equality, we will graph it with a solid line. its slope is ? = −? = ???? ??? = −? ? , and its y-intercept is (?, ?). since the point (?, ?) is easy to calculate with and it is not on the boundary line, we can use it as a test point: ?((?) − ?) + (?) ≤ −? ? ?(−?) ≤ −? ? −? ≤ −? ?... yes so (?, ?) is a solution to ?(? − ?) + ? ≤ −?, and so is every else on the same side of the boundary line as (?, ?). so we indicate this by shading everything on that side of the boundary line. don’t forget to label the equation of the boundary line and the region of solutions. ps 9.1 19) a) find the slope of the line that passes through the points (−?, ?) and (?, ?). ????? = ? = ?? − ?? ?? − ?? = ? − ? ? − (−?) = − ? ? b) produce an equation in slope-intercept form for the line that passes through these two points. (verify that your equation is correct by graphing it in desmos.com) s-i form: ? = ?? + ? substitute the slope in for ?: ? = − ? ? ? + ? and then sub in one of the ordered pairs to find ?: (?) = − ? ? (?) + ? ? = − ? ? + ? ? + ? ? = ? ?? ? = ? so ? = − ? ? ? + ?? ? c) interval:="" (−="" ,="" )="" graph:="" ia="" 1.6="" (no="" systems)="" 17)="" graph="" the="" linear="" inequality="" in="" two="" variables="" by="" hand:="" (?="" −="" )="" +="" ≤="" −?="" solve="" the="" inequality="" for="" :="" (?="" −="" )="" +="" ≤="" −?="" −="" +="" ≤="" −?="" ≤="" −??="" +="" so="" the="" equation="" for="" the="" boundary="" line="" is="" =="" −??="" +="" and,="" since="" the="" inequality="" ≤="" allows="" for="" equality,="" we="" will="" graph="" it="" with="" a="" solid="" line.="" its="" slope="" is="" =="" −?="????" =="" −?="" ,="" and="" its="" y-intercept="" is="" (?,="" ).="" since="" the="" point="" (?,="" )="" is="" easy="" to="" calculate="" with="" and="" it="" is="" not="" on="" the="" boundary="" line,="" we="" can="" use="" it="" as="" a="" test="" point:="" ((?)="" −="" )="" +="" (?)="" ≤="" −?="" (−?)="" ≤="" −?="" −?="" ≤="" −?="" ...="" yes="" so="" (?,="" )="" is="" a="" solution="" to="" (?="" −="" )="" +="" ≤="" −?,="" and="" so="" is="" every="" else="" on="" the="" same="" side="" of="" the="" boundary="" line="" as="" (?,="" ).="" so="" we="" indicate="" this="" by="" shading="" everything="" on="" that="" side="" of="" the="" boundary="" line.="" don’t="" forget="" to="" label="" the="" equation="" of="" the="" boundary="" line="" and="" the="" region="" of="" solutions.="" ps="" 9.1="" 19)="" a)="" find="" the="" slope="" of="" the="" line="" that="" passes="" through="" the="" points="" (−?,="" )="" and="" (?,="" ).="" =="" =="" −="" −="" =="" −="" −="" (−?)="−" b)="" produce="" an="" equation="" in="" slope-intercept="" form="" for="" the="" line="" that="" passes="" through="" these="" two="" points.="" (verify="" that="" your="" equation="" is="" correct="" by="" graphing="" it="" in="" desmos.com)="" s-i="" form:="" =="" +="" substitute="" the="" slope="" in="" for="" :="" =="" −="" +="" and="" then="" sub="" in="" one="" of="" the="" ordered="" pairs="" to="" find="" :="" (?)="−" (?)="" +="" =="" −="" +="" +="" =="" =="" so="" =="" −="" +=""> ? interval: (− ?? ? , ?) graph: ia 1.6 (no systems) 17) graph the linear inequality in two variables by hand: ?(? − ?) + ? ≤ −? solve the inequality for ?: ?(? − ?) + ? ≤ −? ?? − ? + ? ≤ −? ? ≤ −?? + ? so the equation for the boundary line is ? = −?? + ? and, since the inequality ≤ allows for equality, we will graph it with a solid line. its slope is ? = −? = ???? ??? = −? ? , and its y-intercept is (?, ?). since the point (?, ?) is easy to calculate with and it is not on the boundary line, we can use it as a test point: ?((?) − ?) + (?) ≤ −? ? ?(−?) ≤ −? ? −? ≤ −? ?... yes so (?, ?) is a solution to ?(? − ?) + ? ≤ −?, and so is every else on the same side of the boundary line as (?, ?). so we indicate this by shading everything on that side of the boundary line. don’t forget to label the equation of the boundary line and the region of solutions. ps 9.1 19) a) find the slope of the line that passes through the points (−?, ?) and (?, ?). ????? = ? = ?? − ?? ?? − ?? = ? − ? ? − (−?) = − ? ? b) produce an equation in slope-intercept form for the line that passes through these two points. (verify that your equation is correct by graphing it in desmos.com) s-i form: ? = ?? + ? substitute the slope in for ?: ? = − ? ? ? + ? and then sub in one of the ordered pairs to find ?: (?) = − ? ? (?) + ? ? = − ? ? + ? ? + ? ? = ? ?? ? = ? so ? = − ? ? ? + ?? ? c)>