Math 116 week 4 answers

MAT 116 Final Exam - Chapter 0-2 and 6-8

NAME: ___________________________________________

1. List all the factors of 45...

The factors of 45 are = 5,9,3,15,45,1

2. Find the prime factorization of 93.

The prime factors of 93 are 31,3,1 as93 = 31*3*1

3. Find the LCM for 4 and 20.The prime factors of 4 = 2*2*1The prime factors of 20=2*2*5*1The L.C.M = 2*2*5*1=20

A) 2 B) 4 C) 20 D) 80

4.Write the fraction in simplest form.

We can solve this by = 3 / 12 = 1/4

5.Multiply.

5 1/3 = 16/31 ¼ = 5/4

So we get = 16/3 * 5 /4 = 20/3

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6.Divide. ÷

= 77/9

= 7/3

Dividing both we will do like

77/9÷ 7/3 = 77/9*3/7 = 11/3

7.Change the fraction to a decimal.

We need to divide it to change it into decimal

The decimal is 0.46

8. Add. 1.44 + 4.8743

4.8743+1.4400 6.3143

A) 5.3143 B) 6.3143 C) 5.0183 D) 5.9183

9. Evaluate. | 6 | + | –15 |. | 6 | =6 | –15 | = 15So when we add 6 + 15 = 21

A) –21 B) 9 C) –9 D) 21

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10. Which of the following is not an expression?

A) 4(y + 1) B) x + 2y C) ab = 6 D) a + b + c

11. Which property of the real numbers is illustrated by the following statement?9 · (7 + 11) = 9 · 7 + 9 · 11

A) Commutative property of multiplicationB) Associative property of additionC) Associative property of multiplicationD) Distributive property

12. Use the indicated property to write an expression that is equivalent to the following expression.

0 · 1 (commutative property of multiplication)

0.1 *1 = 1*0.1 used the commutative property of multiplicationThe commutative property is a*b = b*a

13.Subtract.

= - 8/8 = -1

14. Subtract. 14 – (–4)14 – (–4) = 14 + 4 = 18

A) 18 B) 10 C) –10 D) –18

15. Determine the range.

11, 7, 3, 21, 22

The difference between the lowest and the highest number is called rangeSo we will subtract 22 and 3 = 22 – 3 = 19

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16. Find for the following set of numbers:

–2, 3, –3, 6, –5, 8, –4

The sum of all the numbers = –2+3+( –3)+6+( –5)+ 8+( –4) =3

A) 6 B) 2 C) 3 D) –1

17. List the like terms in the following group of terms.2b, 2ab2, –3a2b, 4ab2, –4ab2, 6a2

The like term pairs are2ab2 , 4ab2, –4ab2

18. Multiply.b7 b5

the exponential property used in this is so we will get b7 b5

=

19. Divide.

=

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20. Solve. 4(7x + 10) = 3(9x – 5) + 20We will solve like-:4*7x + 4*10= 3*9x –3* 5 + 2028x + 40 = 27x -15 +2028x -27x +40 = -15 +20 subtract 27x on both sides X = -15 + 20 -40 subtracting 40 on both the sides = -35

A) –35 B) 5 C) –5 D) 35

21. Find the mean and the median.–8, –6, 4, 8, 14, 18The mean is = sum of all the numbers total numbers of observations

= –8+ (–6)+4+8+ 14+ 18 = 30 / 6 6

Now for finding the median–8, –6, 4, 8, 14, 18The median is the middle value, so I'll have to rewrite the list in order:–8, –6, 4, 8, 14, 18Total number of observation is even = 6So the formula for finding the median is mean of (n/2 )th and (n/2 + 1)th term.3rd and the 4th term = (4 + 8 )/2 = 12 / 2 = 6

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22. Solve. 10x – 4 – 5x = 6 + 3x – 8

10x – 4 – 5x = 6 + 3x – 8 5x – 4 = 3x – 2 5x – 3x - 4 = -2 subtract 3x on both the sides 2x -4 = -2 2x = -2 +4 2x = 2 add 4 on both sides X = 1

23.Solve. – 6 = 12

– 6 = 12

= 12 + 6 add 6 on both sides

= 18

X = 18 * 5/2 = 45

A) 36 B) C) D) 45

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24.Solve: – =

– =

25. Solve. 10(–x – 7) + 20 = –5(2x + 4)10*(–x) –10* 7 + 20 = –5*2x -5* 4-10x – 70 +20 = -10x -20 -10x +10x = 0Hence we cannot solve this equation and find the solution for x

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26. Of 25 students in a class, 25 passed. What percentage of the students passed?

The number of students in class = 25Number of students passed = 25

Formula to find percentage will become = Number of students passed * 100 The number of students in class = 25/25 *100 = 100 %

So 100% of the students will be passed

27. Fill in the blank with < or >. –11 ___<___ 3

28. An employee who produces x units per hour earns an hourly wage of y = 0.55x + 12 (in dollars). Find the hourly wage for an employee who produces 20 units per hour.hourly wage of y = 0.55x + 12we need to find the hourly wage when x = 20so, y = 0.55x + 12 y = 0.55*20 + 12 = 11.00 +12 =$ 23

A) $23.00 B) $23.50 C) $22.70 D) $23.40

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29. Give the coordinates of the points graphed below.

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30. Graph 3x + y = 3.To graph the linear equation 3x + y = 3 Y= 3 -3x subtract 3x on both sides X 0 1 -1Y 3 0 6

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31. Find the slope of the graphed line.

According to the ordered pairs given we can choose two pairs to find the slope

= (2,0)

= (3,2)

The slope of the equation will be = (2 -0 )/(3-2) = 2/1 = 2

So the slope is 2 (A)

A) 2 B) –2 C) D) –

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32. The following pie chart represents the eye color of students in a certain class.

Blue33%

Brow n50%

Other17%

What percentage of students in the class had blue eyes?

The percentage of students in the class had blue eyes is given as 33%.

33. Find the slope and the y-intercept.y = –5x – 4

to find the y intercept we will put x=0 into the equation y = –5x – 4y = –5x – 4y = -5*0 – 4 = -4

So the y intercept is (0,-4)

And the slope is indicated as -5 as the equation is written in the standard for of y = mx +bso, the slope is -5 and the y intercept is -4

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34.Write the equation of the line with slope and y-intercept (0, 3). Then graph

the line.The equation will be in the form of y = mx +by = mx +bso we put m = -1/2 and y intercept or b = 3

y = -1/2 x + 3 graph the line we will put the convenient values of x to obtain the different values of ya) put x= 0y = -1/2 *0+ 3 = 3The solution is (0,3)b) put x = 2y = -1/2 *2 + 3 = -1 +3 = 2The solution is (2,2)c) put x = -2y = -1/2 *(-2) + 3 = 1 + 3 = 4The solution is (-2,4)X 0 2 -2Y 3 2 4

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35. Write the equation of the line that passes through point (0, –9) with a slope of 6.The equation will be in the form of y= mx + b Put m = 6 and y intercept = -9So, y = 6x - 9A) y = –9x + 6 B) –9x + 6y = 0 C) y = 6x – 9 D) 6y = –9

36. Graph the inequality.x ≤2

Put x = 0, into the equation x ≤20 ≤2Which is true we will shade the portion towards the origin

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37. Given f(x) = –5x – 1, find f(a – 1).f(a – 1) = -5(a- 1) -1 = -5*a +5 -1 = -5a +4A) –5a + 4 B) –5a – 2 C) a – 7 D) a + 4

38. Solve the system by addition.x + y = 6x – y = 4

to solve these equations, x + y = 6 (i)

x – y = 4 (ii) 2x = 10 X = 5Put the value of x into (i)X + y = 65 + y = 6 Y = 6 -5 = 1The values of (x, y) = (5,1)

39. Solve the system by substitution.x + y = 12 y = 2x

to solve these equationsx + y = 12 (i)y = 2x (ii)put y = 2x into the (i) equationwe will get x + y = 12 x + 2x = 12 3x = 12 X = 4

Put this value into the (ii) equation , we will getY = 2 * 4 = 8

The solution is (4,8)

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40. Solve the following system of linear inequalities by graphing.x + 2y 32x – 3y 6

a) x + 2y 3x y subtract 2y on both the sidesputting different values of y to get values of x

Y 0 1 -1X 3 1 5

2x – 3y 6 2x y add 3y on both sides X y

putting different values of y to get values of xY 0 2 -2X 3 6 0

These are the graphs of these two equations but we need to check which area to be shaded, so we will put y = 0,x = 0 in both the equations,0 + 2*0 3 , 0 3 true 2*0 – 3*0 6, 0 6 true

So we will shade the portion including the origin and in between the lines

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A) C)

B) D)

41. Find the GCF for 14 and 21.To find the GCF Find the prime factors of 14 = 2* 7 * 121 = 7 *3*1The common number = 7So the GCF is 7A) 1 B) 6 C) 7 D) 42

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42. Solve the system by addition.2x + 4y = 34x – 2y = 1

2x + 4y = 3 (i)4x – 2y = 1 (ii)

Multiply the first equation (i) by (- 2) so,-4x - 8y = -6 4 x -2 y = 1

-10y = -5 Y = -5/-10 = 1/2Put this value into any equation say (i)2x + 4y = 3 2x + 4 * ½ = 32x + 2 = 3 2x = 3-22x = 1X = 1/2 So the solution is (x,y) = (1/2,1/2)

43. Solve the system by addition or substitution.20x – 5y = –12y = 4x + 3

20x – 5y = –12 (i)y = 4x + 3 (ii)

put the value of (ii) into the (i) equation 20x – 5y = –12 20x – 5*(4x+3) = –12 20x – 20x – 15 = -12 0 – 15 = -12

Thus these two equations are same and cannot have any solution

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44. The difference of two numbers is 33. The second is 7 less than 3 times the first. What are the two numbers?Let the two numbers be x and ySo difference will be ( x – y ) = 33Also given that,The second is 7 less than 3 times the first So ,y = 3x – 7 (i) X –y = 33 (ii) Put the value of y into (ii) equationX – (3x – 7 ) = 33X – 3x +7 = 33 -2x +7 = 33 -2x = 33 -7 -2x = 26 X = 26/-2 = -13Second number will be find by putting the value of x into the (i) equationy = 3x – 7 = 3*(-13) -7= -39 -7 = -46

the two numbers are -46,-13

45. The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 58 in., find the width of the rectangle.

Let the width(w) of the rectangle be xThe length (l)of the rectangle = 2x + 8We are given perimeter = 2(l + w) = 2(2x+8 + x ) = 2(3x + 8) = 6x + 16So 6x + 16 = 58 6x = 58 – 16 6x = 42 X = 7

So the width of the rectangle = 7 in Length = 2*7+8 = 14+ 8 = 22 in

A) 6 in. B) 7 in. C) 8 in. D) 9 in.

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Math 116 week 4 answers

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