Problem solving in context (SAT). Game on.

Problem solving in context (SAT) is a grade 9 math skill aligned to Common Core standard HSS.IC.B.4: use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 problem solving in context (sat) problems our math games drill.

CCSS HSS.IC.B.410 questions in the bank
Kickoff

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Warm-upeasy

A librarian checked 4040 books selected at random from a collection of 800800 books. She found that 1414 of the books in the sample were fiction. Based on the sample, which of the following is the best estimate of the number of fiction books in the collection?

The sample proportion is 1440=720\dfrac{14}{40} = \dfrac{7}{20}. The best estimate for the collection is 720×800=280\dfrac{7}{20} \times 800 = 280 books.

Mid-gameeasy

A ball is tossed upward from a platform. The equation h=4.9t2+14t+21h = -4.9t^2 + 14t + 21 represents this situation, where hh is the height of the ball above the ground, in meters, tt seconds after it is tossed. According to the equation, what is the height, in meters, of the platform from which the ball was tossed?

At the moment of release, t=0t = 0. Substituting gives h=4.9(0)2+14(0)+21=21h = -4.9(0)^2 + 14(0) + 21 = 21 meters.

Mid-gameeasy

A stone is thrown upward from the edge of a cliff overlooking a valley. The equation s=5t2+11t+16s = -5t^2 + 11t + 16 models the stone's height ss, in meters, above the valley floor tt seconds after it is thrown. According to the equation, what is the height, in meters, of the cliff edge above the valley floor?

When the stone is thrown from the cliff edge, t=0t = 0. Then s=5(0)2+11(0)+16=16s = -5(0)^2 + 11(0) + 16 = 16 meters above the valley floor.

Mid-gameeasy

A football is kicked from a stadium deck. The equation d=16t2+48t+64d = -16t^2 + 48t + 64 represents this situation, where dd is the height of the football above the field, in feet, tt seconds after it is kicked. According to the equation, what is the height, in feet, of the deck from which the football was kicked?

When the kick occurs, t=0t = 0. Substituting gives d=16(0)2+48(0)+64=64d = -16(0)^2 + 48(0) + 64 = 64 feet.

Mid-gamemedium

A grocer surveyed 2525 customers selected at random and found that 1515 of them purchased at least one organic item. Based on the survey, which of the following is the best estimate of the number of customers out of 500500 who would purchase at least one organic item?

The sample proportion is 1525=35\dfrac{15}{25} = \dfrac{3}{5}. The best estimate for 500500 customers is 35×500=300\dfrac{3}{5} \times 500 = 300 customers.

Mid-gamemedium

The mass of a chemical sample is modeled by m(x)=480(0.92)xm(x)=480(0.92)^x, where xx is the number of days after an initial measurement. According to the model, by what percent does the mass of the sample decrease each day?

The factor 0.920.92 means that each day the mass is 92%92\% of the previous day's mass, so the daily decrease is 100%92%=8%100\% - 92\% = 8\%.

Mid-gamemedium

An online store's weekly visitors are modeled by v(w)=1,250(1.08)wv(w)=1{,}250(1.08)^w, where ww is the number of weeks after a launch campaign begins. According to the model, how many visitors does the store have 22 weeks after the campaign begins?

Substitute w=2w=2: v(2)=1,250(1.08)2=1,250(1.1664)=1,458v(2)=1{,}250(1.08)^2=1{,}250(1.1664)=1{,}458.

Buzzer beatermedium

A cold front is expected to reduce a reservoir's algae volume by 34\frac{3}{4} of its previous amount each week. The volume is 640640 cubic meters at the start of week 00. Which equation models the algae volume A(w)A(w), in cubic meters, after ww weeks?

A reduction of 34\frac{3}{4} of the previous volume leaves a remaining factor of 14\frac{1}{4} each week. Starting from 640640, the volume after ww weeks is A(w)=640(14)wA(w)=640\left(\frac{1}{4}\right)^w.

Overtime

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