Leaving Certificate Ordinary Level Maths Solutions ...

Leaving Certificate Ordinary Level Maths Solutions SEC 2013 Sample Paper 1 1. (a)

.

6−2

0.027

2.8%

812

900%

( √3 )4

220

1.05 × 106

24 × 25

1 _

(b)

8·85 8.9 8.95 9.0 9.05 9.1 9.15 9.2 9.25 9.3 9.35 9.4 9.45 9.5 8.9

2. (a)

__

9

37 4

82

F = P(1 + i)t F = 5000(1.06)8 F = €7969.24

(b)

€100 after 8 years gives €120 F = P(1 + i)t 120 = 100(1.0r)8 1.2 = (1.0r)8 8

____

1.0r = √ 1.2

1.0r = 1.023 0.0r = 0.023 r = 2.3% 3. (a)

z1 = (3 + 2i)(2 − 5i) z1 = 6 − 15i + 4i − 10i2 z1 = 6 − 15i + 4i + 10 z1 = 16 − 11i z 2 = (5 + 4i)(17 − 13i) − (5 + 3i)(17 − 13i) z2 = (17 − 13i)[(5 + 4i) − (5 + 3i)] z2 = (17 − 13i)[5 + 4i − 5 − 3i] z2 = (17 − 13i)(i) z2 = 17i − 13i2 z2 = 13 + 17i

1

9.3 × 100

3p

) (

(

5 __ 7 5 __ 1 __ z3 = __ 2 + 2i − 2 + 2i 5 __ 7 __ 5 __ 1 z3 = __ 2 + 2i − 2 − 2i 6 z3 = __ 2i z3 = 3i

)

z4 = 1 + i + i2 + i3 z4 = 1 + i − 1 − i z4 = 0 (b)

z1 = 16 − 11i

z2 = 13 + 17i

|z1| = √162 + (−11)2

|z2| = √132 + 172

|z1| = √377

|z2| = √458

|z 1| = 19.416

|z2| = 21.4

___________ ____

________ ____

∴ z2 is further from origin than z1. 4. (a)

x2 − 6x − 23 = 0

_______

− b ± √b − 4ac x = _____________ 2

2a a = 1, b = − 6, c = − 23

_______________

−(−6) ± √(−6)2 − 4(1)(−23) _______________________ x= 2(1) _______ 6 ± √36 + 92 x = ___________ 2 ____ 6 ± √128 x = ________ 2______ 6 ± √64 × 2 x = __________ 2__ 6 ± 8√2 x = _______ 2 __

x = 3 ± 4√2 (b)

2r − s = 10 rs − s2 = 12 s = 2r − 10 ⇒ r(2r − 10) − (2r − 10)2 = 12

When r = 7

When r = 8

⇒ 2r − 10r − (4r + 100 − 40r) = 12

2r − s = 10

2r − s = 10

⇒ −2r2 + 30r − 100 = 12

2(7) − s = 10

2(8) − s = 10

∴ r2 − 15r + 56 = 0

⇒s=4

⇒s=6

2

2

(r − 7)(r − 8) = 0 ⇒r=7

or

r=8 2

5. (a)

(b)

1 21

Pattern No. of tiles

2 33

3 45

4 57

5 69

Tn = a + (n − 1)d Tn = 21 + (n − 1)12 Tn = 12n + 9

(c)

Tn = 12n + 9 T10 = 12(10) + 9 T10 = 129

(d)

Tn = 12n + 9 = 399 12n = 390 n = 32.5 Pattern 32

(e)

(f)

n Sn = __ 2 [2a + (n − 1)d ] n Sn = __ 2 [2(21) + (n − 1)(12)] n Sn = __ 2 [42 + 12n − 12] n Sn = __ 2 [30 + 12n] Sn = n[30 + 12n] 21 + 33 + 45 + 57 + 69 + 81 + 93 = 399 7 patterns

6. (a)

(b)

Day Plant 1 (height, cm) Plant 2 (height, cm)

0 16 24

1 20 27.5

Height of plant = h, Start height = s,

2 24 31

3 28 34.5

4 32 38

Number of days = d,

5 36 41.5

Growth rate = r

Plant 1: h = r(d) + s

h = 4d + 16

Plant 2: h = r(d) + s

h = 3.5d + 24

3

6 40 45

7 44 48.5

(c)

Height (cm) 120 100 80 60 40 20 0

Days 0 2

4

6

8

10 12 14 16 18 20 22 24 26 28 30

Answer: (16, 80)

(d)

(i)

(e)

h = 4d + 16 and h = 3.5d + 24 ⇒ 4d + 16 = 3.5d + 24 ⇒ 0.5d = 8 ∴ d = 16 h = 4(16) + 16 h = 80 ∴ The point of intersection is (16, 80)

(f)

It is a more accurate answer.

(g)

John is assuming that, after the first seven days, the plants grow at the same rate each day.

7. (a)

h(x) = x2 + 1 = 50 ⇒ x2 = 49

___

∴ x = √49 = ±7 x=7 (b)

(i)

or x = −7 x g(x)

0 −0.5

1 −1

1.5 −4

1.75 2.25 −2 4

4

2.5 2

3 1

4 0.5

(ii)

y

4 3 2 1 0

0

–1

1

–1

2

3

–2 –3 –4

(c)

(i)

5 f (x) = x − __ x ⇒ f (x) = x − 5x−1 f ′(x) = 1 + 5x−2 5 f ′(x) = 1 + __2 x

(ii)

5 Slope of tangent is f ′(x) = 1 + __2 x Slope of y = 6x is 6 As they are parallel 5 ⇒ 1 + __2 = 6 x 5 __ =5 x2 1 __ =1 x2 ⇒ x2 = 1 __

x = √1 = ±1 x=1

or

5 y = 1 − __ 1 y = −4

x = −1 5 y = −1 − ___ −1 y=4

Points are (1, −4) and ( −1, 4) 8. (a) (b)

y = x2 − 6x + 1 (i)

dy ___ = 2x − 6 dx

f(x) = 5 − 3x f(x + h) = 5 − 3(x + h) f (x + h) − f (x) f ′(x) = lim ____________ h h→0 5 − 3(x + h) ] − [ 5 − 3x ] [ f ′(x) = lim ____________________ h h→0 −3h f ′(x) = lim ____ h→0 h f ′(x) = lim(−3) = −3 h→0

5

4

x

(ii)

y = (x2 − 4)(3x − 1) u v dy dv du ___ = u___ + v ___ dx dx dx dy ___ = (x2 − 4)(3) + (3x − 1)(2x) dx dy ___ = 3x2 − 12 + 6x2 − 2x dx dy ___ = 9x2 − 2x − 12 dx dy When x = 2 ___ = 9(2)2 − 2(2) − 12 dx = 36 − 16 = 20

(c)

(i)

v = 96 + 40t − 4t2 v = 96 ⇒ 96 + 40t − 4t2 = 96 ∴ 40t − 4t2 = 0 t(40 − 4t) = 0 t = 0 sec

(ii)

or

t = 10 sec

dv Acceleration (a) = ___ = 40 − 8t dt at t = 2.5 sec Acceleration (a) = 40 − 8(2.5) = 20 ms−2

(iii)

Acceleration (a) is negative, ⇒ a < 0 a = 40 − 8t ⇒ 40 − 8t < 0 8t > 40 t > 5 sec

6

Leaving Certificate Ordinary Level Maths Solutions SEC 2013 Sample Paper 2 1. (a)

If one task can be accomplished in x different ways, and following this another task can be accomplished in y different ways, then the first task followed by the second can be accomplished in x by y different ways.

(b)

5! ways = 120 ways

(c)

(5!)(3!)(2!) ways = 1,440 ways

2. (a)

E(X) = 1(0.25) + 2(0.25) + 3(0.15) + 4(0.15) + 5(0.1) + 6(0.1) = 2.9

(b)

E(X) = 1(0.25) + 2(0.25) + 3(0.15) + 4(0.15) + 5(0.1) + 6(0.1) = 2.9 E(X) = 1(0.166) + 2(0.166) + 3(0.166) + 4(0.166) + 5(0.166) + 6(0.166) = 3.5 If you play the game many times with a fair die, you will win an average of €0.50 per game, but if you play with the biased die, you will lose an average of €0.10 per game.

3. (a)

(b)

y A (3,5) 5 4 3 B (–6,2) 2 1 0 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x –1 –2 –3 C –4 (4,–4) –5

(3, 5)

(−6, 2)

(x1, y1)

(x2, y2)

y2 − y1 ______ −3 __ 2 − 5 ___ 1 Slope of AB: ______ x2 − x1 = −6 − 3 = −9 = 3 Equation of AB: y − y1 = m(x − x1) 1 y − 5 = __ (x − 3) 3 3(y − 5) _____ x−3 _______ = 3 3 3y − 15 = x − 3 x − 3y + 12 = 0

7

(c)

METHOD 1 (3, 5)

METHOD 2

→ (0, 0) 6

(− 6, 2) → (−9, −3)

A

4

(4, −4) → (1, −9)

B 2 9 units –6

4 –4

2 –2

2

4

6

–2

(−9, −3)

(1, − 9)

(x1, y1)

(x2, y2)

4 –4 10 units –6

C

1 Area of Δ: __[x1 y2 − x2 y1] 2 1 __ [ (−9)(−9) − (1)(−3) ]

Area of triangle ABC = Area of rectangle – Area of shaded triangles = 90 – 48 = 42 units2

2

1 __ [81 + 3] 2

42 units2 4. (a) 5 c

4 3 2 Q(3, 1)

1 0 0 1 –8 –7 –6 –5 –4 –3 –2 –1 –1 P(–2, –1) –2 –3 –4 –5 –6 –7

8

2

3

4

5

(b)

( −2, −1)

(3, 1)

(x , y )

(x , y )

1 1 2 2 __________________

Radius: √(x2 − x1)2 + ( y2 − y1)2 _______________

(3 + 2)2 + (1 + 1)2 √________ (5)2 + 22 √______ √25 + 4 ___

√29

Centre (h, k):

___

Radius r = √29 ___

Centre (−2, −1): Radius √29

Equation of c: (x − h)2 + (y − k)2 = r2

(c)

Equation of c: (x + 2)2 + (y + 1)2 = 29 y2 − y1 _____ 1 + 1 __ 2 Slope of PQ: P(−2, − 1), Q(3, 1) ______ x2 − x 1 = 3 + 2 = 5 y2 − y1 _____ 6−1 5 __ Slope of QR: Q(3, 1), R(1, 6) ______ x2 − x1 = 1 − 3 = − 2 Opposite slopes ∴ PQ ⊥ QR ∴ QR is a tangent to c.

5.

4.2 cm

5.5 cm

4.5 cm

3.3 cm

3.7 cm 0 cm

A 3.1 cm

3.1 cm

3.1 cm

3.1 cm B

3.1 cm

15.5 cm

h A ≈ __ [ y1 + yn + 2(y2 + y3 + y4 + ... + yn−1) ] 2 3.1 A ≈ ___[ 4.2 + 0 + 2(5.5 + 4.5 + 3.3 + 3.7) ] 2 A ≈ 59.21 cm2 6A.

(a)

The converse of a theorm is formed by taking the conclusion as a starting point and having the starting point as a conclusion.

(b)

Statment: All squares are rectangles. Converse (false): All rectangles are not squares.

9

6B.

|∠ABQ| = ( 180° − (a + g ) ) ..........three angles in ΔABQ |∠DCB| = (180° − a)...........opposite angle of cyclic quadrilateral

A

⇒|∠BCP| = [180° − (180° − a)] = a.............straight angle

a

|∠ABQ| = (a + b )..........exterior angle of the ΔBCP ⇒( 180° − (a + g ) ) = (a + b )

B

⇒ 180° − a − g = a + b ⇒ 180° − a − a = b + g ⇒ b + g = 180° − 2a Q

The event that, on average, takes the longest to complete is the CYCLE.

(ii)

In all three histograms, the times are grouped into intervals of 2 MINUTES.

(iii)

The time of the fastest person in the swim was between 28 AND 30 MINUTES.

(iv)

The median time for the run is approximately 25 MINUTES.

(v)

The event in which the times are most spread out is the CYCLE.

(i)

B.

(ii)

40 Frank Run time (minutes)

(b)

(i)

35 30

(iii)

6 people took longer than Frank to compete the cycle

25 20 15 30

35

40 45 50 55 Cycle time (minutes)

60

40 Run time (minutes)

(a)

C g

Q.E.D.

7.

b

D

35 30 25 20 15 30

Brian

35

40 45 50 55 Cycle time (minutes)

10

60

Brian’s times would be very unusual because, as we can see from the scatter graph, the correlation between his run time and cycle time seem to be out of line with the rest of the athletes. His run time seem very fast compared to his cycle time.

P

(c)

(i)

1. There seems to be a greater spread in the times among the female 30 –39-yearolds compared with the male 40 – 49-year-olds. 2. The median time for the males appears to be less than the median time for the females. 3. The distribution of times for the males is approximately normal, whereas the distribution for the female times is more skewed.

(ii)

This diagram might cause Máire to change her belief, as it seems from the diagram that the males have a better mean time.

8. (i)

C C

20 cm 25 cm

22 cm

18 cm

D 60°

E

A

F

C

25 22 (iii) ______ = _________ sin 60° sin|∠AFC| 25 22 ______ = _________ 0.8660 sin|∠AFC| 25(0.8660) sin |∠AFC| = _________ 22 25(0.8660) |∠AFC| = sin–1 _________ 22 |∠AFC| = 79.77° ∴ |∠ACF| = 40.23° (iv)

25 cm

22 cm

60° A

F

C

20 cm

40.23° 18 cm

|DE|2 = 202 + 182 – 2(20)(18) cos 40.23° |DE| = 13.20 cm

D E

11

9. (a)

Six boxes: each box is 0.3 m, so height of man = 1.8 m

(c) 240 cm 90 cm

270 cm

180 cm

150 cm

Six squares ≈ 180 cm, which is the average height of a man One square ≈ 30 cm or 0.3 m By counting the square, we can get approximate measurements. The approximate measurements are in the diagram above. To find the approximate volume, we use these approximate measurements. Volume of hopper = volume of cylinder + volume of top cone + volume of bottom cone 1 1 = p r2h + __ p r2h + __ p r2h 3 3 1 1 = p (1.2)2(2.7) + __ p (1.2)2(0.9) + __ p (1.2)2(1.5) 3 3 = p (1.2)2[2.7 + 0.3 + 0.5] = p 1.44[3.5] = p (5.04) = 15.12 m3 (taking p ≈ 3)

12

Leaving Certificate Ordinary Level Maths Solutions Sample Paper 1, no. 1 1. (a)

(i)

5

(ii)

3

(b)

125

(c)

√5 = 52

(d)

125 __ 5x = ____ √5

__

1 _

53 5x = ___1 52 1 _

5x = 522 1

x = 2_2 2. (a)

2x2 + 11x + 5 (2x + 1)(x + 5) ∴ l = 2x + 1

(b)

2x2 + 11x +5 = 0 (2x + 1)(x + 5) = 0 2x + 1 = 0 1 x = –__ 2

(c)

or x + 5 = 0 or

x = –5

a(b – c) = d ab – ac = d

(d)

ab = ac + d ac + d b = ______ a 3 3 1 a = __, c = __, d = __ 5 2 4

( ) ( )

3 3 1 _ _ _ 2 4 + 5 ________ b= 1 _ 2 19

b = 1__ 20

or 1.95

13

3. (a) (b)

Re(z) = –3, Im(z) = 2i _

z = –3 – 2i _________

(c)

_____

___

|z| = √(–3)2 + 22 = √9 + 4 = √13 ___________

_

_____

___

|z| = √(–3)2 + (–2)2 = √9 + 4 = √13 (d)

Im 3i z

2i |z| i Re

–3

–2

0 1 –i

–1 – |z|

2

3

–2i

– z

–3i

4. (a) (b) (c)

(d) 5. (a)

D = S × T = 480 × 3 = 1,440 km D 1,440 S = __ = _____ = 360 km/hr T 4 D 2,800 T = __ = _____ = 5.6 hours 500 S = 5 hrs 36 minutes 1,440 + 2,800 S = ____________ = 493.02 km/hr 3 + 5.6 Week

1

2

3

4

5

6

David

8

10

12

14

16

18

Ella

3

5.5

8

10.5

13

15.5

14

(b)

y

(6,18)

18 16

(6,15.5)

14

David

12 10 Ella 8

(1,8)

6 4 (1,3)

2 0 –2

(c)

–1 –2

x 0

1

2

3

4

5

6

7

8

9

10

The graphs are both linear.

(d)

(11,28)

y 35 30 (11,28)

25 20 15

David

10 Ella

5 0

x 0

2

4

6

8

6

12

14

16

–5

(e)

In Week 11 of their training programme, both runners will be running 28 km.

(f)

David: Tn = 8 + (n – 1)(2) Tn = 8 + 2n – 2 Tn = 6 + 2n Ella:

Tn = 3 + (n – 1)(2.5) Tn = 3 + 2.5n – 2.5 Tn = 0.5 + 2.5n

(g)

David: T20 = 6 + 2(20) = 46 km Ella:

T20 = 0.5 + 2.5(20) = 50.5 km

15

(h)

Multiple possible answers. Example: David: Currently, David will reach 42 km after 18 weeks, so he could repeat Week 9 and Week 18 of his training programme. Alternatively, he could delay increasing his distance for the first two weeks. Ella: At the moment, Ella will reach 42 km during her 16th week of the programme. She could repeat Weeks 4, 8, 12 and 16 of her programme.

6. (a)

(i)

x2 + y2 = 25

x+y=7

(7 – y)2 + y2 = 25

x=7–y

49 – 14y2 + y2 + y2 – 25 = 0 2y2 – 14y + 24 = 0 y2 – 7y + 12 = 0 (y – 3)(y – 4) = 0 y=3 x=7–3

or

x=7–4

x=4

x=3

(4,3)

(3,4) _________________

(ii)

d = √(x2 – x1)2 + (y2 – y1)2 _______________

= √(3 – 4)2 + (4 – 3)2 _________

= √(–1) + (1)2 __

= √2

__

Width of tunnel: √2 × 10 = 14.14 metres (b)

(i)

A – Graph 3 B – Graph 2 C – Graph 1

(ii)

A ⇒ –1, 7 B ⇒ 2, 2 C ⇒ –2, 4

16

or y = 4

2x2 – 10x – 3 = 0

(iii)

x2 + x2 – 10x – 3 = 0 x2 – 10x – 3 = –x2 x2 – 4x – 3 = –x2 + 6x x2 – 4x + 4 = –x2 + 6x + 7 B=A Solution: –0.2, 5.2 7. (a)

(i)

__

__

f ( √3 ) = √3 2 + 1 = 4 g(1) = 2(1) = 2

(ii)

4 = k(2) 2=k 2

(b)

(i)

x +2 _____ 4 – x2 dy _______________________ (4 – x2)(2x) – (x2 + 2)(–2x) ___ = dx (4 – x2)2 8x – 2x3 + 2x3 + 4x = ________________ (4 – x2)2 12x = _______ (4 – x2)2

(ii)

y = (3 – 2x2)5 dy ___ = 5(3 – 2x2)4 (– 4x) dx When x = 1 ⇒ 5[3 – 2(1)2]4 (– 4(1)) = 5[1]4 (–4) = –20

(c)

h = 400 – 16t2 dh ___ = –32t dt

(height) (speed)

2

dh ___ = –32 dt2

(acceleration/deceleration)

17

(i)

h=0 400 – 16t2 = 0 400 = 16t2 400 2 ____ =t 16 20 ___ s=t 4 5s = t

(ii)

Height at t = 0 h = 400 – 16(0)2 h = 400 m

(iii)

Speed = –32t = –32(5) = –160 m/s Yes, it will withstand the impact.

8. (a)

f (x) = 2x3 – 3x2 – 12x + 4 f (–2.5) = 2(–2.5)3 – 3(–2.5)2 – 12(–2.5) + 4 = –16 f (–1) = 2(–1)3 – 3(–1)2 – 12(–1) + 4 = 11 f (1) = 2(1)2 – 3(1)2 – 12(1) + 4 = –9 f (3) = 2(3)2 – 3(3) – 12(3) + 4 = –5

(b)

At y-axis, x = 0 ∴ y=4

(0,4)

(c)

f ′(x) = 6x2 – 6x – 12

(d)

6x2 – 6x – 12 = 0 at local max/min x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2, x = –1 y = 2(2)3 – 3(2)2 – 12(2) + 4 = –16 Min (2,–16)

Max (–1,11) 18

(e)

y (–1,11)

10 5 (0,4) x

0 –4 –3.5 –3 –2.5 –2 –1.5 –1 –0.5

0 0.5 1 1.5 2 2.5 3 3.5 4

–5

(3,–5) (1,–9)

–10 2x3 – 3x2 – 12x + 4 –15 (–2.5,–16)

(f)

f (x) = 0

(g)

–1 < x < 2

⇒

(2,–16)

x = –2, 0.3 in the domain given.

19

Leaving Certificate Ordinary Level Maths Solutions Sample Paper 1, no. 2 1. (a) (b) (c)

A prime number is a natural number that has only two factors, 1 and itself. 2, 3, 5, 7, 11, 13, 17, 19 2 2 3 3

36 18 9 3 1

2 3 3 7

36 = 22 × 32 (d)

126 63 21 7 1

126 = 2 × 32 × 7

HCF = 2 × 3 × 3 = 18 LCM = 2 × 3 × 3 × 2 × 7 = 252

(e) 2. (a)

362,013 = (22 × 32)2,013 = 24,026 × 34,026 x 5 5 E = {x | __ − __ ≤ __, x ∈ Z } 2 3 6 5 5 x __ __ − ≤ __ 2 3 6 1 2 3x − 10 ≤ 5

3

4

5

6

7

8

–5 –4 –3 –2 –1

0

1

2

3x ≤ 15 x≤5 (b)

F = {x | 4 + 3x > − 2, x ∈ Z } 4 + 3x > −2 3x > −6 x > −2

(c)

E∩F –3 –2 –1

(d) 3. (i) (ii)

0

1

2

3

4

5

6

G ∩ E: {1, 2, 3, 4, 5} x + 3 cm 3 1 1 __ x (x + 3) = __x2 + __x cm2 2

2

2

20

(iii)

(

1 x2 + 4 __ x(x + 3) 2 2 x + 2x2 + 6x

)

3x2 + 6x cm2 3x2 + 6x = 105

(iv)

3x2 + 6x − 105 = 0 x2 + 2x − 35 = 0 (x − 5)(x + 7) = 0 x − 5 = 0 or

x+7=0

x = 5 or

x = −7

x ≠ −7 as length is not negative ∴ x = 5 cm 4. (i)

Im 4i 3i 2i

w

i |w| Re –4

–3

–2

0 1 2 |z + –i w|

–1

–2i

3

4 z+w

|z|

–3i

z

(ii)

z + w = 2 − 3i + 1 + 2i = 3 − i

(iii)

|z + w| is the diagonal of the parallelogram 0w(z + w)z

(iv)

|w| = √12 + 22 = √5

______

__

_________

___

|z| = √22 + (−3)2 = √13 _________

___

|z + w| = √32 + (−1)2 = √10 5. (i) (ii)

Arithmetic Tn = 10 + (n − 1)(5) = 10 + 5n − 5 = 5n + 5

21

(iii)

5n + 5 = 450 5n = 445 n = 89 weeks

(iv)

(v)

n Sn = S89 = __{2a + (n − 1)d} 2 89 = ___{2(10) + (89 – 1)(5)} 2 89 = ___{20 + 440} 2 89 = ___{460} 2 = €20,470 Week 1 2 3 4 5 6

Donations 560,000 280,000 140,000 70,000 35,000 17,500

(vi) 700000€ Donations 600000€

(1,560000)

500000€ 400000€ 300000€

(2,280000)

200000€

(3,140000) (4,70000) (5,35000) (6,17500)

100000€ 0€ 0

0.5

1

1.5

2

2.5

3

3.5

4

()

19 (vii) T10 = 560,000 __ = €1,093.75 2 1 6. (a) 2 ± __(0.025) 2 = 2 ± 0.0125 = (1.9875, 2.02125) (b)

A – Reject B – Accept C – Accept D – Accept 22

4.5

5

5.5

6

6.5

Week Number 7

7.5 8

(c)

B: 2.025 mg/L C: 1.95 mg/L D: 2.05 mg/L

(d)

0.035 Sample A: % error = _____ × 100 1.85 = 1.89% 0.0225 Sample B: % error = ______ × 100 2.025 = 1.11% 0.045 Sample C: % error = _____ × 100 1.95 = 2.31% 0.045 Sample D: % error = _____ × 100 2.05 = 2.20 %

(e)

A – Reject B – Reject C – Reject D – Reject

(f)

2.013 × (2.205 × 10−6) × 0.22 = 0.0000009765063 lb/gallons

(g) 7. (a)

9.765063 × 10−7 lb/gallons 3x − 12 < 0 3x < 12 x