Subject Tests Answer Explanations Math

7/25/2019 Subject Tests Answer Explanations Math

1/10

MathematicsLevel 1 and Level 2

SAT Subject Tests

Answer Explanations to Practice Questionsfrom Getting Ready for the SAT Subject Tests

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the Subject Tests in Mathematics

Level 1 and Level 2.

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2/10

261.80 255.90= 0 5. 9

20 10

n250+ 5 9. 0

10= 250+0 5. 9n

SAT

Subject Tests in Mathematics Level 1 and Level 2

n this document, you will find detailed answer explanations to all of the mathematics

practice questions from Getting Ready for the SAT Subject Tests. y rev ew ng t ese

answer explanations, you can amiliarize yoursel with the types o questions on the

test and learn your strengths and weaknesses. The estimated di culty level is based

on a scale o 1 to 5, with 1 the easiest and 5 the most di cult.

o find out more about the SAT Subject Tests, visit us at SATSubjectTests.org.

SAT

Subject Test inMathematics Level 1

lineAC is a transversal to the parallel linesAB and CD,

the alternate interior angles BDC and DBAare o equal

measure. Tus, the measure o BDC isy, which is also

the measure o EDC. Te measure o DCE is given to

be z . Tereore, the sum o the angle measures o EDC,

in degrees, is (180 x) +y + z. Te triangle sum theorem

applied to EDC gives the equation (180 x) +y + z = 180,

which can be solved or xto arrive at x =y + z.

1. Difficulty: 2

Choice (B) is correct.

Te cost o the equipment is $250, and each package o

10 blank CDs costs $5.90. Te total cost or the band toproduce 10 CDs, that is, 1 package, is 250 + 5.90 = 255.90

dollars, and the total cost to produce 20 CDs, that is, 2

packages, is 250 + 5.90(2) = 250 +11.80 = 261.80 dollars.

Alternatively, one can apply the interior angle sum t rem

heo

to pentagonABECD. Since lineAD is a transversal to the

parallel linesAB and CD, it ollows that BADand ADC

are supplementary; that is, the sum o the measures o

these two angles is 180. Te measure o BEC, interior

to polygonABECD, is (360 x). Te measure o EBA is

given to be y , and the measure o DCEis given to be z .

Tereore, the sum o the measures o the interior angles o

pentagonABECD is 180 + (360 x) +y + z. Te interior

angle sum theorem applied to pentagonABECD gives the

equation 180 + (360 x) +y+ z= 540, which can be solved

or xto arrive at x =y + z.

One way to determine the correct expression is to nd

the slope-intercept orm o the equation o the line that

passes through the two points given by the ordered

pairs (10, 255.90) and (20, 261.80). Te line has slope

andy-intercept 250, which is

the cost, in dollars, or the band to produce 0 CDs afer

purchasing the equipment. Te equation o this line is

= 0.59n + 250y , where n .is the number o CDs used

Alternatively, note that the total cost to produce 10 CDs

is 250 +5.90 = 250 +0.59(10) dollars, and the cost to

produce 20 CDs is 250 +5.90(2) =250 +0.59(20) dollars,

so, in general, i nis a multiple o

produce n CDs is ( )10, then the total cost to

dollars.

3. Difficulty: 2

Choice (C) is correct.

One way to determine the value o nis to create and

solve an algebraic equation. Te phrase a number

nis increased by 8 is represented by the expression

n ++ 8, and the cube root o that result is equal to 0.5, so3n + =8 0.5. Solving or n gives n + =8 ( .0 5)3 = 0.125,

and so n = 0.125 =8 8.125.

2. Difficulty: 2

Choice (A) is correct.

Let E be the point o intersection o linesACand BD.B

C

yzx

E

DA

Alternatively, one can invert the operations that were done

to n. Apply the inverse o each operation, in the reverse

order: First cube 0.5 to get 0.125, and then decrease thisvalue by 8 to nd that n = 0.125 =8 8.125.

One way to determine xin terms oyand zis to nd the

measure o each o the angles o EDCin terms o x,yand

z and then apply the triangle sum theorem. Since CED

is supplementary to BECand the measure o BEC is

given to be x, the measure o CED is (180 x).Since

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3/10

60= 1.

405,

a b+ 9

b = 4.

< 3 3 < 2 > 2

( +3) Negative Positive Positive

( 2) Negative Negative Positive

( + 3) ( 2) Positive Negative Positive

SAT


4. Difficulty: 3


o determine the value o , apply the act that two complex

numbers are equal i and only i the real and pure imaginary

parts are equal. Since (a b) 5+ i 9 ,+ = + ai this gives the

two equations = and 5i ai;= that is, a b 9and+ = 5

= a b 9,, 5 + = and. Tereore

5. Difficulty: 3

Cho ce (C) s correct.

One way to determine all values of x for

which 4 x2 x 2 is rst to rewrite the inequality inan equivalent form that compares a factored expression

to 0 and then reason about the arithmetic sign of the24 x x 2product of the factors. The inequality

q valent to x24 x+ 2 0,is e ui which in turn isequ x2 x+ 6 .ivalent to 0 Since x2 +x 6 factors

as2( )1 (x + x 6) = ( 1 () x+ 3 () x 2)

inequality is equivalent to (1)(x +3)(x 2) 0 or

(x + 3)(x 2) 0. To solve (x +3)(x 2) 0, notice

that this inequality is satised by a value of x precisely

when either x = 3, x =2 or the product of the factors

(x + 3) and (x 2) is negative; this last condition is true

for 3 < x


4/10

x2 9x 3

x2 9x 3

x 9 (x + 3)(x 3)= = x 3

x 3 x 3

1= 1

1

BC,

2 2 2316t + 46t + = 16 t t + 55

20 8 = 12.

SAT


8. Difficulty: 3

Choice (E) is correct.

Statement I is alse: Te unctionfhas the real line as its

domain. Te unctionghas as its domain all values o x

exceppt 3, because the value o the expression is

undened when 3 is substituted or x. Tis means that the

graph of contains a point with x-coordinate equal to 3, butthe graph ogcontains no such point.

Statement II is true: For any x not equal to 3, the

expressions x + 3and give the same value, since or

2

x 3, the expression + . It

ollows that or x 3, the graphs of andg contain the same

point with that x-coordinate.

Statement III is true: Since statement II is true, or every

number xnot equal to 3, the point (x, x +3) is on both

the graph ofand the graph og. Since there are innitelymany numbers xnot equal to 3, the graphs have an innite

number o points in common.

9. Difficulty: 4

Choice (D) is correct.

Since line C is perpendicular to the line segment with

endpoints (2, 0) and (0, 2), the slope o line C must be

the negative reciprocal o the slope o the line segment. Te0 ( )2

= 12 0

line segment has slope , so the slope o line C

equals .

10. Difficulty: 3Choice (B) is correct.

It is given that 3 students have sampled all three kinds o

candy bars and that 5 students have sampled exactly two

kinds, so a total o 3 +5 = 8 students have sampled more

than one kind o candy bar. Since 20 students have sampled

one or more o the three kinds o candy bars, the number o

students that have sampled only one kind is

1. Difficulty: 3


o determine the length o side apply the denition o

the tangent o an angle in a right triangle: the tangent ois equal to the length o the side opposite angle divided

by the length o the si e a jacent to angle . Tus, in thisBC BC

case, tan22 = = , which gives BC = 10tan22.AC 10

Use a calculator, set to degree mode, to compute that

10tan22 4.0.

12. Difficulty: 4


One way to determine the maximum height the ball reaches

is to rewrite the quadratic expression that denes the

unction by completing the square:

( )82 23 23

2 23 2= 16(t t + ( ) ( ) ) + 58 16 16

2 223 23= 16((t ) ( ) ) + 516 162 223 23

= 16((t ) + 16( ) + 516 16It is not necessary to simpliy any urther, as the maximum

23height must correspond to t=

16, which is the only value

232

o tthat makes the term 16(t 16) nonnegative. By2

23 23= 16 5 38.0625. Tereore, tosubstitution, h + =(16) (16)

the nearest oot, the maximum height the ball reaches is

38 eet.

Alternatively, one can use a graphing calculator

to determine the maximum value o the unction

. Since h( ) = 5 1, h( ) = +6 46 5+ = 350 1 ,

h( ) = 16 4 ( ) + 5= 33and2 ( ) + 462

h( ) 1 ( ) + 46 3( ) + 5 = 1, the maximum value o3 = 6 9

must occur or some t-value between 1 and 2. Set the

window so that the independent variable goes rom 0 to

3 and the dependent variable goes rom 0 to 50 to view

the vertex o the parabola. Upon tracing the graph, the

maximum value o is slightly greater than 38. Tereore,

to the nearest oot, the maximum height the ball reaches is

38 eet.

13. Difficulty:4


One way to determine the volume o the solid is to

determine the length C, width w, and height o the solid,

in centimeters, and then apply the ormula V = Cwh to

compute the volume. Let C, , and represent the length,

width and height, in centimeters, respectively, o the solid.

Te area o the ront ace o the solid is Ch = 24square

centimeters, the area o the side ace is w = 8 square

centimeters, and the area o the bottom ace is Cw = 3square centimeters. Elimination o by using the rst

Ch 24 Ctwo equations gives = , which simplies to = 3,wh 8 w

or C= 3w. Substitution o 3 or C in the third equation

gives (3w w 3, or 3 3, so 1 (since only positive

values o make sense as measurements o the length

o any edge o a rectangular solid). Substitution o 1 or

in the equation w 8 gives = 8, and substitution o

8 or in the equation Ch = 24gives 8C= 24, so C= 3.

Tereore, the volume V o the solid, in cubic centimeters, is

V = ( )( )( )3 1 8 = 24.



5/10

SAT


Alternatively, one can recognize that the square o the

volume o a rectangular solid is the product o the areas

o the ront, side and bottom aces o the solid. Tat

is, squaring both sides o the ormulaV = Cwh gives

wh w = ( )w hV 2 = C C h C ( )C (wh). Tereore, in this case,

V 2 = ( )(3 24 8)( ) = 576, so V = 576 = 24. Note that it is

not necessary to solve or the values o C , and .

4. Difficulty: 4


Te area o the shaded region can be ound by subtracting

the area o rectangleABCDrom the area o the circle.

o determine the area o the circle, rst nd the radius

, and then compute the area . Since rectangleAB D

is inscribed in the circle, ABCis an inscribed right

angle, and thus AC is a diameter o the circle. Applying

the Pythagorean theorem to right triangleAB , one

nds he length o side AC is 52 + 122 = 169 = 13.

Tus, the radius o the circle is13

, and the area o the2 2

13 169

circle is = . Te area o rectangleAB D is( 2) 4 =169

5 12 60and thereore, the area o the shaded region is

60 72.7.4

15. Difficulty: 3

Cho ce (E) s correct.

To determ ne how many real numbers sat s y

f (k) = 2 is to determine how many solutions the4 3 2equat on x 3x 9x + = has, wh ch n turn4 2

s to determ ne how many solut ons the equat on4 3 2x 3x 9x 2 0+ = has. It is not difcult to see, using

4 3 2the Rat onal Roots Theorem, that x 3x 9x + 2 as

no factor of the form or any whole or rat onal

num er va ue o k, but there are rea num er va ues o

such that is a factor. This problem must be solved

us ng a nonalgebra c method.

One way to determ ne how many numbers sat s y

f k( ) = 2 s to exam ne the graph of the funct on4 3 2f x( ) = x 3x 9x + 4. Use a graph ng calculator

4 3 2to graph y = x 3x 9x + 4for xon a su tablylarge nterval to see all ntersect ons of the graph w th

the l ne = 2, and then count the number of po nts of

intersection.

y

50

O

50

55x

There are at least four such po nts: A rst po nt w th

x-coordinate between 2 and 1, a second point with-coor nate etween 1 and 0, a th rd po nt w th

x-coord nate between 0 and 1 and a fourth po nt

with -coordinate between 4 and 5. The fact that4 3 2x 3x 9x + 4 s a polynom al of degree 4 means

that there can be at most four such po nts. Therefore,

t ere are our va ues o k for wh ch ( ) = 2.

Alternat vely, one can exam ne a table of values4 3 2of the funct on f x( ) = x 3x 9x + 4 an t en

dent fy ntervals for wh ch the values of ( ) = 2

at the endpo nts have d fferent s gns and apply the

Intermed ate Value Theorem to each of those ntervals.4 3 2Create a ta e o va ues or

y=

x 3

x 9

x+ 4 or

w o e num er va ues o etween and 5, nclus ve,

and count the number of ntervals of length 1 for wh ch

t e va ue o s greater t an 2 or one o t e en po nts

and less than 2 for the other endpo nt.

x f(x)

5 779

4 308

3 85

2 8

1 1

0 4

1 7

2 40

3 77

4 76

5 29



6/10

x= 10,000.

SAT


There are four such ntervals: [ , , 1, 0], [0, 1] an

[4, 5]. By the Intermed ate Value Theorem, for each of

these four intervals, there is a value n that nterval

suc t at ( ) 2. Th s shows that there are at least four4 3 2suc va ues o , an t e act t at x 3x 9x + 4 s a

polynom al of degree 4 means that there can be at most

our suc va ues o k. Therefore, there are four values of

or w c ( ) 2.

6. Difficulty: 5


o determine the value, to the nearest hundred dollars, o the

automobile when t= 4, nd the equation o the least-squares

regression line and then evaluate that equation at t 4.

A calculator can be used to nd the least-squares

regression line. Te specic steps to be ollowed depend

on the model o calculator, but they can be summarized

as ollows. Enter the statistics mode, edit the list o

ordered pairs to include only the our points given in the

table and perorm the linear regression. Te coefficientswill be, approximately, 15,332 or the -intercept and

2,429 or the slope, so the regression line is given by

the equationy = 2 429t+ 15 332. When t=4, one gets, ,

y= 2 429 4, ( ) + 15 332, = 5 616. Tereore, based on the,

least-squares linear regression, the value, to the nearest

hundred dollars, o the automobile when t=4 is $5,600.

SAT

Subject Test inMathematics Level 2

121(3x+ 12 ) 3 +

xis equivalent to the expression x =12

. As1

(2x 12 ) 2 x x

12gets innitely larger, the expression approaches the

x12

3 +xvalue 0, so as xgets innitely larger, the expression

12

2

x3 0 3+

approaches the value = . Tus, as gets innitely2 0 2

larger, ) approaches3

.2

Alternatively, one can use a graphing calculator to estimate

the height o the horizontal asymptote or the unction3x+ 12 3x+ 12

f x( ) = . Graph the unction y =12

on an2x 12 2x

interval with large x-values, say, rom x= 100 to x= 1,000.

y

2.5

2.0

1.5

1.0

.5

O 100 200 300 400 500 600 700 800 900 1,000x

By examining the graph, they-values all seem very close

to 1.5. Graph the unction again, rom, say, x= 1,000 to

y

2.5

2.0

1.5

1.0

.5

xO 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000

7. Difficulty: 2


Te distance dbetween the points with coordinates

( , ,x y z ) and ( , ,x y z ) is given by the distance ormula:1 1 1 2 2 22 2 2d = (x x ) + (y y ) + (z z ) .2 1 2 1 2 1

Tereore, the distance between the points

with coordinates ( , , 7) and (2, 1, 4) is:

, which simplies to 83 9 11..

8. Difficulty: 2


One way to etermine the value that ) approaches

as gets innitely larger is to rewrite the denition o

the unction to use only negative powers o and then

reason about the behavior o negative powers o xas x ets

innitely larger. Since the question is only concerned with3x+ 12

what happens to as gets innitely larger, one can2x 12

3x+ 12assume that is positive. For 0, the expression

2x 12

Te -values vary even less rom 1.5. In act, to the scale

o the coordinate plane shown, the graph o the unction3x+ 12

f x =2 2

is nearly indistinguishable rom the( )x 1

asymptotic liney=1.5. Tis suggests that as xgets innitely

larger, ) approaches 1.5, that is, 3.2

Note: Te algebraic method is preerable, as it provides3

a proo that guarantees that the value ) approaches is .2

Although the graphical method worked in this case, it

does not provide a complete justication; or example,

the graphical method does not ensure that the graph

resembles a horizontal line or very large -values such as100

101

10 x 10 .



7/10

SAT


9. Difficulty: 4


According to the model, the worlds population in

January 1995 was 5.3(1.02)5 and in January 1996 was

5.3(1.02) . Tereore, according to the model, the

population growth rom January 1995 to January 1996,

in billions, was 5.3(1.02) 5.3(1.02)5, or equivalently,

.5

.9

117 000,0005 3 1 02( . ) ( 0 02)(10 ) ,

0. Difficulty: 4


First, note that the angle o the parallelogram with vertex

(3, 5) is one o the two larger angles o the parallelogram:

Looking at the graph o the parallelogram in the -plane

makes this apparent. Alternatively, the sides o the angle

o the parallelogram with vertex (3, 5) are a horizontal line

segment with endpoints (3, 5) and (6, 5), and a line segment

o positive slope with endpoints (2, 1) and (3, 5) that

intersects the horizontal line segment at its lef endpoint

(3, 5), so the angle must measure more than 90. Since thesum o the measures o the our angles o a parallelogram

equals 360, the angle with vertex (3, 5) must be one o the

larger angles.

One way to determine the measure o the angle o

the parallelogram with vertex (3, 5) is to apply the

Law o Cosines to the triangle with vertices (2, 1),

(3, 5) and (6, 5). Te length o the two sides o the angle2 2with vertex (3, 5) are (3 2) (5 1) = 17+

2 2and (3 6) + ( =5 5) 3; the length o the side2 2opposite the angle is (6 2) (5 1) = 4 2 .+

Let represent the angle with vertex (3, 5) and apply2 2 2the Law o Cosines: A +B 2ABcos=C, so

2

2 2C (A + B ) 32 (17 + 9) 6 1cos= = = =

2AB 2( 17 )( ) 6 17 171.

3

Tereore, the measure o one o the larger angles o the

parallelogram is arccos( 1 ) 104 0. .17Another way to etermine the measure o the angle o the

parallelogram with vertex (3, 5) is to consider the triangle

(2, 1), (3, 5) and (3, 1). Te measure o the angle o this

triangle with vertex (3, 5) is 90less than the measure o

the angle o the parallelogram with vertex (3, 5). Te angle

o the triangle has opposite side o length 3 =2 1 and

adjacent side o length 5 1 4 = , so the measure o this

angle is arctan1

. Tereore, the measure o the angle o the4

parallelogram with vertex (3, 5) is 90 + arctan1

4 104 0. .

Yet another way to determine the measure o the angle o

the parallelogram with vertex (3, 5) is to use trigonometric

relationships to nd the measure o one o the smaller

angles, and then use the act that each pair o a larger and

smaller angle is a pair o supplementary angles. Consider

the angle o the parallelogram with vertex (2, 1); this angle

coincides with the angle at vertex (2, 1) o the right triangle

with vertices at (2, 1), (3, 5), and (3, 1), with opposite side

o length 5 1 =4 and adjacent side o length 3 2 = 1, so

the measure o this angle is arctan4. Tis angle, together

with the angle o the parallelogram with vertex (3, 5), orm

a pair o interior angles on the same side o a transversal

that intersects parallel lines, so the sum o the measures

o the pair o angles equals 180. Tereore, the measure

o the angle o the parallelogram with vertex (3, 5) is

180 arctan4 104.0.

21. Difficulty:4


o determine the numerical value o the ourth term,

rst determine the value o tand then apply the common

difference.

Since 2t, 5t 1, and 6t+ 2 are the rst three terms

o an arithmetic sequence, it must be true that

(6t+ 2) (5t1) = (5t 1) 2t, that is, t+ 3 = 3t 1.

Solving t+ 3 = 3t 1 or tgives t 2. Substituting 2 or tin

the expressions o the three rst terms o the sequence, one

sees that they are 4, 9 and 14, respectively. Te common

difference between consecutive terms or this arithmetic

sequence is 5 = 14 9 = 9 4, and thereore, the ourth

term is 14 + 5 = 19.

22. Difficulty:3


o determine the height o the cylinder, rst express the

diameter o the cylinder in terms o the height, and then

express the height in terms o the volume o the cylinder.

Te volume o a right circular cylinder is given b

2

y V=

r h,where ris the radius o the circular base o the cylinder

and his the height o the cylinder. Since the diameter1

and height are equal, h = 2r. Tus, r = h. Substitute th2

e1

expression h or2

rin the volume ormula to eliminate r:

1 2

4V = ( h) h = h3 . Solving or h 3gives h = V . Since2 4 the volume o the cylinder is 2, the height o the cylinder is

h = 38

1 3. 7

.

23. Difficulty: 3


One way to determine the value o sin( )

is to apply the sine o difference o two angles

identity: sin( ) = sin cos cossin.

Since sin= 0 and cos= 1, the identity gives

sin( ) = 0(cos) ( 1)(sin ) = sin

sin( ) = 0 5. 7.

Another way to determine the value o sin( ) is to apply

the supplementary angle trigonometric identity or the sine:

sin( ) = sin. Tereore, sin( ) = 0 5. 7.



8/10

SAT


4. Difficulty: 4


One way to determine the probability that neither person

selected will have brown eyes is to count both the number

o ways to choose two people at random rom the people

who do not have brown eyes and the number o ways to

choose two people at random rom all 10 people, and then

compute the ratio o those two numbers.

Since 60 percent o the 10 people have brown eyes, there

are 0.60(10) =6 people with brown eyes, and 10 6 = 4

people who do not have brown eyes. Te number o ways

o choosing two people, neither o whom has brown eyes,

i4 3( )

s = 6: Tere are 4 wa2

ys to choose a rst person and

3 ways to choose a second person, but there are 2 ways in

which that same pair o people could be chosen. Similarly,

the number o ways o choosing two people at random10 9( )

rom the 10 people is = 45 2

. Tereore, the probability

that neither o the two people selected has brown eyes is

6 0 1. 3.45

Another way to determine the probability that neither

person selected will have brown eyes is to multiply the

probability o choosing one o the people who does not

have brown eyes at random rom the 10 people times the

probability o choosing one o the people who does not have

brown eyes at random rom the 9 remaining people afer

one o the people who does not have brown eyes has been

chosen.

Since 60 percent o the 10 people have brown eyes, the

probability o choosing one o the people who does not have

brown eyes at random rom the 10 people is 1 0.60 = 0.40.I one o the people who does not have brown eyes has been

chosen, there remain 3 people who do not have brown eyes

out o a total o 9 people; the probability o choosing one o

the 3 people who does not have brown eyes at random rom3

the 9 people is . Tereore, i two people are to be selected9

rom the group at random, the probability that neither3

person selected will have brown eyes is 0 4. 0( 0 1. 39 ) .5. Difficulty: 2


By the Factor Teorem, x 2 is a actor o

x3 + kx2 + 12x 8 only when 2 is a root o

x3 + kx2 + 12x 8, t at is,( )2 3 + k( )2 2h + 12( )2 =8 0,

which simplies to 4k + 24 =0. Tereore, k = 6.

Alternatively, one can perorm the division o

x3 + kx2 + 12x 8 by x 2 and then nd a value or k so

that the remainder o the division is 0.

Since the remainder is 4 + 4, the value o must satisy4 + 24 = 0. Tereore, = 6.

26. Difficulty: 4


One way to determine the value o f 1( .1 5)is to solve the

equation f x( ) = 1 5. ince ( ) = 3or x. S f x x3 + 1, start with

the equation3

x3 + =1 1.5, and cube both sides to get

x3 + =1 (1 5. )3 = 3.375. Isolate xto get x3 = 2.375, and

apply the cube root to both sides o the equation to get

x = 3 2 375. . 1 3.

Another way to determine the value o

1

f

( .1 5)is to nd aormul f 3a or 1 and then evaluate at 1.5. Let y = x3 + 1

and solve or x: cubing bot y3 = x3h sides gives + 1, so

x3 = y3 1 = 3y3 1. T 1, and x ereore, f ( )y = 3y3 1,

and f 1(1 5. ) = 3 (1 5. )3 1 = 3 2.375 1 3. .

27. Difficulty:4


One way to determine which o the equations best models

the data in the table is to use a calculator that has a statistics

mode to compute an exponential regression or the data.

Te specic steps to be ollowed depend on the model o

calculator but can be summarized as ollows: Enter the

statistics mode, edit the list o ordered pairs to include

only the our points given in the table and perorm an

exponential regression. Te coefficients are, approximately,

3.3 or the constant and 1.4 or the base, which indicates

that the exponential equation = 3.3(1.4)xis the result oy

perorming the exponential regression. I the calculator

reports a correlation, it should be a number that is very

close to 1, which indicates that the data very closely matches

the exponential equation. Tereore, o the given models,

= 3.3(1.4)xbest ts the data.y

Alternatively, without using a calculator that has a statistics

mode, one can reason about the data given in the table.

Te data indicates that as xincreases,yincreases;

thus, options A and B cannot be candidates or such a

relationship. Evaluating options C, D and Eat x = 9.8

shows that option D is the one that gives a value o that isy

closest to 0.12. In the same way, evaluating options C, D and

Eat each o the other given data points shows that option D

is a better model or that one data point than either option

C or option E. Tereore, = 3.3(1.4)xis the best o they

given models or the data.



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t= 91 t= 92 corresponds to the day

t= 91 days afer18 7

= 1.6 5 1

=

.

1f x = +( ) 2

xis the set

1

y= +2xt

,or some x-value.

1f x +( ) =

x2

1y += 2

x

1y= +2

x

y =21

x1

x=y2

.1

x=y2

y= +2,,x

1

1f x( ) = +2

x

1.

x

1

xcan take on any value

1+2

x

1f x( ) = +2

x

35 7 2+ sin t

3 3 365(

2 t =1

365sin ,( ) which corresponds to

2 365t= , so t= = 91.25.

365 2 4However, or this problem,

35 7 2 35 7 2+ sin (91) > sin (92)

3+

3 3 365 3 36535 7 2

+ sin t3 3 365

35 7 23

+3

sin365

(41)

35 7 2 35 7 2

3+

3sin

365(91)

3+

3sin

365(41) 0( ( )) ( ( ))

C= 1.02(20) +93.6373.

t= 0

y

t (5 t) 2,7+ = + + 10 + 30

+ 31 + 20

10 + 30 +1 = 41

SAT


8. Difficulty: 4


Statement I is alse: Since C= 1.02F+93.63, high values o

Fare associated with low values o C, which indicates that

there is a negative correlation between C

and F.

Statement II is true: When 20 percent o calories are rom

at,F=20 and the predicted percent o calories rom

carbohydrates is

Statement III is true: Since the slope o the regression

line is 1.02, as Fincreases by 1, Cincreases by

1.02(1) = 1.02; that is, Cdecreases by 1.02.

9. Difficulty: 3


One way to determine the slope o the line is to compute

two points on the line and then use the slope ormula. Forexample, letting gives the point (5, 7) on the line, and

lettingt= 1 gives the point (6, 8) on the line. IFSFGPSF, the

slope o the line is equal to

Alternatively, one can express in terms o x. Since x=5 + t,

y=7 + t it ollows thaty= x +2.and

Tereore, the slope o the line is 1.

30. Difficulty: 3


Te range o the unction dened by

oy-values such tha

One way to determine the range o the unction dened

by is to solve the equation ,or x

and then determine whichy-values correspond to at least

one x-value. o solve ,or x, rst subtract 2 rom

both sides to get and then take the reciprocal

o both sides to get Te equation

shows that or anyy-value other than 2, there is an x-value

such that and that there is no such x-value or

y= 2. Tereore, the range o the unction dened by

is all real numbers except 2.

Alternatively, one can reason about the possible values

o the term Te expression

except 0, so the expression can take on any value

except 2. Tereore, the range o the unction dened by

is all real numbers except 2.

31. Difficulty:4


o determine how many more daylight hours the day with

the greatest number o hours o daylight has than May 1,

nd the maximum number o daylight hours possible orany day and then subtract rom that the number o daylight

hours or May 1.

o nd the greatest number o daylight hours possible

or any day, notice that the expression )is maximized when

tmust be a whole number, as it represents a count o days

afer March 21. From the shape o the graph o the sine

unction, either or

with the greatest number o hours o daylight, and since

, the expression( ) ( )is maximized when( )

March 21. (It is not required to nd the day on which the

greatest number o hours o daylight occurs, but it is

days afer March 21, that is, June 20.)

Since May 1 is days afer March

21, the number o hours o daylight or May 1 is

( )Tereore, the day with the greatest number o hours o daylight

has


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