If there is one prayer that you should

- Samuel Dominic Chukwuemeka
**pray/sing** every day and every hour, it is the
LORD's prayer (Our FATHER in Heaven prayer)

It is the **most powerful prayer**.
A **pure heart**, a **clean mind**, and a **clear conscience** is necessary for it.

For in GOD we live, and move, and have our being.

- Acts 17:28

The Joy of a Teacher is the Success of his Students.

- Samuel Chukwuemeka

Prerequisite: Introductory Statistics

**Calculators:**

Vertical Data Entry: Measures of Position

Horizontal Data Entry: Measures of Position

For ACT Students

The ACT is a timed exam...$60$ questions for $60$ minutes

This implies that you have to solve each question in one minute.

Some questions will typically take less than a minute a solve.

Some questions will typically take more than a minute to solve.

The goal is to maximize your time. You use the time saved on those questions you
solved in less than a minute, to solve the questions that will take more than a minute.

So, you should try to solve each question __correctly__ and __timely__.

So, it is not just solving a question correctly, but solving it __correctly on time__.

Please ensure you attempt __all ACT questions__.

There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students

Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students

Any question labeled WASCCE is a question for the WASCCE General Mathematics

Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students

__For the Questions:__

Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.

Any comma included in a number indicates a decimal point.

__For the Solutions:__

Decimals are used appropriately rather than commas

Commas are used to separate digits appropriately.

Solve all questions.

Show all work.

(1.) Students $A, B, C, D.\:\:and\:\:E$ had their $SAT$ (Scholastic Assessment Test) scores converted
to $z-scores$.

Their $z-zcores$ are $-2.00, -1.00, 0.00, 1.00,\:\:and\:\:2.00$ respectively.

Who made the highest of the five scores?

Student $E$ made the highest test score because the score is $2.00$ standard deviations above the mean.

Their $z-zcores$ are $-2.00, -1.00, 0.00, 1.00,\:\:and\:\:2.00$ respectively.

Who made the highest of the five scores?

Student $E$ made the highest test score because the score is $2.00$ standard deviations above the mean.

(2.) Two friends Agnes and Agatha were arguing on who made the better score.

Agnes made a $73\%$ on her Chemistry test while Agatha made a $46\%$ on her Biology test.

The Biology test scores have a mean of $65\%$ and a standard deviation of $10\%$, while the Chemistry test scores have a mean of $95\%$ and a standard deviation of $11\%$.

Who made the better score?

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \underline{Agnes'\:\:Chemistry\:\:Test} \\[3ex] x = 73 \\[3ex] \mu = 95 \\[3ex] \sigma = 11 \\[3ex] z = \dfrac{73 - 95}{11} \\[5ex] z = -\dfrac{22}{11} \\[5ex] z = -2.00 \\[3ex] \underline{Agatha's\:\:Biology\:\:Test} \\[3ex] x = 46 \\[3ex] \mu = 65 \\[3ex] \sigma = 10 \\[3ex] z = \dfrac{46 - 65}{10} \\[5ex] z = -\dfrac{19}{10} \\[5ex] z = -1.90 \\[3ex] -1.90 \gt -2.00 \\[3ex] $ Therefore, Agatha's Biology test score is better because it has a higher $z-score$ than Agnes' Chemistry test score.

*
Note the responses of your students. *

If some of them said that Agnes' test score was better, re-teach the concept of $z-scores$

Agnes made a $73\%$ on her Chemistry test while Agatha made a $46\%$ on her Biology test.

The Biology test scores have a mean of $65\%$ and a standard deviation of $10\%$, while the Chemistry test scores have a mean of $95\%$ and a standard deviation of $11\%$.

Who made the better score?

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \underline{Agnes'\:\:Chemistry\:\:Test} \\[3ex] x = 73 \\[3ex] \mu = 95 \\[3ex] \sigma = 11 \\[3ex] z = \dfrac{73 - 95}{11} \\[5ex] z = -\dfrac{22}{11} \\[5ex] z = -2.00 \\[3ex] \underline{Agatha's\:\:Biology\:\:Test} \\[3ex] x = 46 \\[3ex] \mu = 65 \\[3ex] \sigma = 10 \\[3ex] z = \dfrac{46 - 65}{10} \\[5ex] z = -\dfrac{19}{10} \\[5ex] z = -1.90 \\[3ex] -1.90 \gt -2.00 \\[3ex] $ Therefore, Agatha's Biology test score is better because it has a higher $z-score$ than Agnes' Chemistry test score.

If some of them said that Agnes' test score was better, re-teach the concept of $z-scores$

(3.) $IQ$ scores are measured with a test designed so that the mean is $107$ and the standard
deviation is $19$.

(a.) What are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores?

(b.) What are the $IQ$ scores that separate the unusual $IQ$ scores from the usual $IQ$ scores?

(a.) A data score is__usual__ if $-2.00 \le z \le 2.00$

A data value is__unusual__ if the $z-score \lt -2.00$ OR the $z-score \gt 2.00$

The lower $z-score$ boundary is $-2.00$

The upper $z-score$ boundary is $2.00$

This implies that $-2.00$ and $2.00$ are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores

(b.) To find the $IQ$ scores (data values), we need to express them in terms of the $z-scores$.

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \rightarrow x - \mu = z\sigma \\[3ex] x = z\sigma + \mu \\[3ex] Lower\:\:IQ\:\:score\:\:boundary = -2(19) + 107 = -38 + 107 = 69 \\[3ex] Upper\:\:IQ\:\:score\:\:boundary = 2(19) + 107 = 38 + 107 = 145 \\[3ex] Usual\:\:IQ\:\:scores:\:\: 69 \le IQ \le 145 \\[3ex] Unusual\:\:IQ\:\:scores:\:\: IQ \lt 69 \:\:OR\:\: IQ \gt 145 \\[3ex] $ Therefore $69$ and $145$ are the $IQ$ scores that separate the unusual $IQ$ scores from the usual $IQ$ scores

(a.) What are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores?

(b.) What are the $IQ$ scores that separate the unusual $IQ$ scores from the usual $IQ$ scores?

(a.) A data score is

A data value is

The lower $z-score$ boundary is $-2.00$

The upper $z-score$ boundary is $2.00$

This implies that $-2.00$ and $2.00$ are the $z-scores$ that separate the unusual $IQ$ scores from the usual $IQ$ scores

(b.) To find the $IQ$ scores (data values), we need to express them in terms of the $z-scores$.

$ z = \dfrac{x - \mu}{\sigma} \\[5ex] \rightarrow x - \mu = z\sigma \\[3ex] x = z\sigma + \mu \\[3ex] Lower\:\:IQ\:\:score\:\:boundary = -2(19) + 107 = -38 + 107 = 69 \\[3ex] Upper\:\:IQ\:\:score\:\:boundary = 2(19) + 107 = 38 + 107 = 145 \\[3ex] Usual\:\:IQ\:\:scores:\:\: 69 \le IQ \le 145 \\[3ex] Unusual\:\:IQ\:\:scores:\:\: IQ \lt 69 \:\:OR\:\: IQ \gt 145 \\[3ex] $ Therefore $69$ and $145$ are the $IQ$ scores that separate the unusual $IQ$ scores from the usual $IQ$ scores

(4.) At one time, President Barack Obama had a net worth of $\$3,670,505.00$

The $17$ members of the Executive Branch had a mean net worth of $\$4,939,455.00$ with a standard deviation of $\$7,775,948.00$ (OpenSecrets.org)

(a.) What is the difference between the mean net worth of all the Executive Branch members and President Obama's net worth?

(b.) How many standard deviations is the difference?

(c.) What is the $z-score$ of President Obama's net worth?

(d.) Is President Obama's net worth usual or unusual?

$ (a.)\:\: Difference = 4939455 - 3670505 = \$1,268,950.00 \\[3ex] (b.)\:\: x = 4939455 \\[3ex] \mu = 3670505 \\[3ex] \sigma = 7775948 \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{4939455 - 3670505}{7775948} \\[5ex] z = \dfrac{1268950}{7775948} \\[5ex] z = 0.16 \\[3ex] The\:\:difference\:\:is\:\:0.16\:\:standard\:\:deviations \\[3ex] (c.)\:\: \underline{Obama's\:\:worth} \\[3ex] x = 3670505 \\[3ex] \mu = 4939455 \\[3ex] \sigma = 7775948 \\[3ex] z-score = \dfrac{3670505 - 4939455}{7775948} \\[5ex] z-score = -\dfrac{1268950}{7775948} \\[5ex] z-score = - 0.1631891057 \\[3ex] z-score \approx -0.16 \\[3ex] (d.)\:\:Because -2.00 \le -0.16 \le -2.00 \\[3ex] $ Obama's net worth is usual.

The $17$ members of the Executive Branch had a mean net worth of $\$4,939,455.00$ with a standard deviation of $\$7,775,948.00$ (OpenSecrets.org)

(a.) What is the difference between the mean net worth of all the Executive Branch members and President Obama's net worth?

(b.) How many standard deviations is the difference?

(c.) What is the $z-score$ of President Obama's net worth?

(d.) Is President Obama's net worth usual or unusual?

$ (a.)\:\: Difference = 4939455 - 3670505 = \$1,268,950.00 \\[3ex] (b.)\:\: x = 4939455 \\[3ex] \mu = 3670505 \\[3ex] \sigma = 7775948 \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{4939455 - 3670505}{7775948} \\[5ex] z = \dfrac{1268950}{7775948} \\[5ex] z = 0.16 \\[3ex] The\:\:difference\:\:is\:\:0.16\:\:standard\:\:deviations \\[3ex] (c.)\:\: \underline{Obama's\:\:worth} \\[3ex] x = 3670505 \\[3ex] \mu = 4939455 \\[3ex] \sigma = 7775948 \\[3ex] z-score = \dfrac{3670505 - 4939455}{7775948} \\[5ex] z-score = -\dfrac{1268950}{7775948} \\[5ex] z-score = - 0.1631891057 \\[3ex] z-score \approx -0.16 \\[3ex] (d.)\:\:Because -2.00 \le -0.16 \le -2.00 \\[3ex] $ Obama's net worth is usual.

(5.) **HSC Mathematics Standard 1** The heights, in centimetres, of 10 players on a basketball team are shown.

Justify your answer with calculations.

Fences: Lower Fence and Upper Fence determine outliers of data.

A data value is an outlier if it is less than the lower fence or greater than the upper fence.

Because the question asked for the height of the shortest player, we shall be concerned with only the Lower Fence.

Data is already sorted

$ n = 10 \\[3ex] \dfrac{1}{4} * 10 \\[5ex] = 2.5th \approx 3rd\;\;position \\[3ex] Q_1 = 185 \\[5ex] \dfrac{3}{4} * 10 \\[5ex] = 7.5th \approx 8th\;\;position \\[3ex] Q_3 = 194 \\[5ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 194 - 185 \\[3ex] IQR = 9 \\[3ex] LF = Q_1 - 1.5(IQR) \\[3ex] LF = 185 - 1.5(9) \\[3ex] LF = 185 - 13.5 \\[3ex] LF = 171.5 \\[3ex] $ Because the height of the shortest player is 170 cm and 170 < 171.5; the height of the shortest player is an outlier.

170, 180, 185, 188, 192, 193, 193, 194, 196, 202

Is the height of the shortest player on the team considered an outlier?Justify your answer with calculations.

Fences: Lower Fence and Upper Fence determine outliers of data.

A data value is an outlier if it is less than the lower fence or greater than the upper fence.

Because the question asked for the height of the shortest player, we shall be concerned with only the Lower Fence.

Data is already sorted

$ n = 10 \\[3ex] \dfrac{1}{4} * 10 \\[5ex] = 2.5th \approx 3rd\;\;position \\[3ex] Q_1 = 185 \\[5ex] \dfrac{3}{4} * 10 \\[5ex] = 7.5th \approx 8th\;\;position \\[3ex] Q_3 = 194 \\[5ex] IQR = Q_3 - Q_1 \\[3ex] IQR = 194 - 185 \\[3ex] IQR = 9 \\[3ex] LF = Q_1 - 1.5(IQR) \\[3ex] LF = 185 - 1.5(9) \\[3ex] LF = 185 - 13.5 \\[3ex] LF = 171.5 \\[3ex] $ Because the height of the shortest player is 170 cm and 170 < 171.5; the height of the shortest player is an outlier.

(6.) **WASSCE** In a community of $500$ people, the $75th$ percentile age is $65$ years while the $25th$ percentile
age is $15$ years.

How many of the people are between $15$ and $65$ years?

$ n = 500 \\[1em] \underline{65\:years} \\[1em] 75th\:\:percentile = \dfrac{75}{100} * 500 = 75 * 5 = 375 \\[2em] \underline{15\:years} \\[1em] 25th\:\:percentile = \dfrac{25}{100} * 500 = 25 * 5 = 125 \\[2em] \underline{Between\:15\:years\:\:and\:\:65\:years} \\[1em] 375 - 125 = 250\:people $

How many of the people are between $15$ and $65$ years?

$ n = 500 \\[1em] \underline{65\:years} \\[1em] 75th\:\:percentile = \dfrac{75}{100} * 500 = 75 * 5 = 375 \\[2em] \underline{15\:years} \\[1em] 25th\:\:percentile = \dfrac{25}{100} * 500 = 25 * 5 = 125 \\[2em] \underline{Between\:15\:years\:\:and\:\:65\:years} \\[1em] 375 - 125 = 250\:people $

(7.)

Let us represent this information in a table.

$ GPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] GPA = \dfrac{55}{16} \\[5ex] GPA = 3.4375 \\[3ex] GPA \approx 3.44 $

Let us represent this information in a table.

Course | Grade | Point | Credit Hour (CH) | Grade Point, (GP) = Point * Credit Hour |

Algebra | $A$ | $4$ | $4$ | $4 * 4 = 16$ |

English | $B$ | $3$ | $3$ | $3 * 3 = 9$ |

Biology | $B$ | $3$ | $4$ | $3 * 4 = 12$ |

Chemistry | $A$ | $4$ | $4$ | $4 * 4 = 16$ |

Chinese | $C$ | $2$ | $1$ | $2 * 1 = 2$ |

$\Sigma CH = 16$ | $\Sigma GP = 55$ |

$ GPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] GPA = \dfrac{55}{16} \\[5ex] GPA = 3.4375 \\[3ex] GPA \approx 3.44 $

(8.)

$ \underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50 $

$ \underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50 $

(9.)

Based on the solution in that link, $m = 3$

Based on the solution in that link, $m = 3$

(10.)

$ \Sigma f = 3 + 1 + 3 + 4 + 9 = 20 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 3 = 7 \\[3ex] 7 + 4 = 11...stop \\[3ex] $ The class that contains the median is $81-85$

$ \Sigma f = 3 + 1 + 3 + 4 + 9 = 20 \\[3ex] \dfrac{\Sigma f}{2} = \dfrac{20}{2} = 10 \\[5ex] Begin\:\:from\:\:the\:\:first\:\:class \\[3ex] Keep\:\:adding\:\:the\:\:frequencies\:\:till\:\:you\:\:get\:\:to\:\:10 \\[3ex] 3 + 1 = 4 \\[3ex] 4 + 3 = 7 \\[3ex] 7 + 4 = 11...stop \\[3ex] $ The class that contains the median is $81-85$

(11.)

$ \Sigma x = 5 + 8 + x + 12 + (x + 5) + 10 \\[3ex] \Sigma x = 35 + x + x + 5 \\[3ex] \Sigma x = 40 + 2x \\[3ex] n = 6 \\[3ex] \bar{x} = 10 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] 40 + 2x = 10(6) \\[3ex] 40 + 2x = 60 \\[3ex] 2x = 60 - 40 \\[3ex] 2x = 20 \\[3ex] x = \dfrac{20}{2} \\[5ex] x = 10 \\[3ex] \underline{Check} \\[3ex] x = 10 \\[3ex] x + 5 = 10 + 5 = 15 \\[3ex] Data\:\:values = 5, 8, 10, 12, 15, 10 \\[3ex] \Sigma x = 5 + 8 + 10 + 12 + 15 + 10 = 60 \\[3ex] n = 6 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{60}{6} \\[5ex] \bar{x} = 10 $

$ \Sigma x = 5 + 8 + x + 12 + (x + 5) + 10 \\[3ex] \Sigma x = 35 + x + x + 5 \\[3ex] \Sigma x = 40 + 2x \\[3ex] n = 6 \\[3ex] \bar{x} = 10 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] 40 + 2x = 10(6) \\[3ex] 40 + 2x = 60 \\[3ex] 2x = 60 - 40 \\[3ex] 2x = 20 \\[3ex] x = \dfrac{20}{2} \\[5ex] x = 10 \\[3ex] \underline{Check} \\[3ex] x = 10 \\[3ex] x + 5 = 10 + 5 = 15 \\[3ex] Data\:\:values = 5, 8, 10, 12, 15, 10 \\[3ex] \Sigma x = 5 + 8 + 10 + 12 + 15 + 10 = 60 \\[3ex] n = 6 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{60}{6} \\[5ex] \bar{x} = 10 $

(12.)

Let the unreadable weight = $x$

$ For\:\:20\:\:patients \\[3ex] \bar{x} = 82 \\[3ex] n = 20 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 82(20) = 1640 \\[3ex] For\:\:19\:\:patients \\[3ex] \bar{x} = 86 \\[3ex] n = 19 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 86(19) = 1634 \\[3ex] \Sigma x_{19} + x = \Sigma x_{20} \\[3ex] 1634 + x = 1640 \\[3ex] x = 1640 - 1634 \\[3ex] x = 6 $

Let the unreadable weight = $x$

$ For\:\:20\:\:patients \\[3ex] \bar{x} = 82 \\[3ex] n = 20 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \rightarrow \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 82(20) = 1640 \\[3ex] For\:\:19\:\:patients \\[3ex] \bar{x} = 86 \\[3ex] n = 19 \\[3ex] \Sigma x = \bar{x} * n \\[3ex] \Sigma x = 86(19) = 1634 \\[3ex] \Sigma x_{19} + x = \Sigma x_{20} \\[3ex] 1634 + x = 1640 \\[3ex] x = 1640 - 1634 \\[3ex] x = 6 $

(13.)

$ Pass\:\:mark = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:4 = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:3 = 1 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:2 = 3 \\[3ex] Number\:\:of\:\:students\:\:who\:\:failed\:\:the\:\:test = 5 + 1 + 3 = 9 $

$ Pass\:\:mark = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:4 = 5 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:3 = 1 \\[3ex] Number\:\:of\:\:students\:\:who\:\:made\:\:2 = 3 \\[3ex] Number\:\:of\:\:students\:\:who\:\:failed\:\:the\:\:test = 5 + 1 + 3 = 9 $

(14.)

There is no mode.

Data Value | Frequency |

$51$ | $1$ |

$61$ | $1$ |

$62$ | $1$ |

$57$ | $1$ |

$50$ | $1$ |

$67$ | $1$ |

$68$ | $1$ |

$58$ | $1$ |

$53$ | $1$ |

There is no mode.

(25.)

$ 2(4x + 1) = 8 - 3(x - 5) \\[3ex] 8x + 2 = 8 - 3x + 15 \\[3ex] 8x + 3x = 8 + 15 - 2 \\[3ex] 11x = 21 \\[3ex] x = \dfrac{21}{11} \\[5ex] $__Check__

$ 2(4x + 1) = 8 - 3(x - 5) \\[3ex] 8x + 2 = 8 - 3x + 15 \\[3ex] 8x + 3x = 8 + 15 - 2 \\[3ex] 11x = 21 \\[3ex] x = \dfrac{21}{11} \\[5ex] $

$ \underline{LHS} \\[3ex] 2(4x + 1) \\[3ex] x = \dfrac{21}{11} \\[5ex] 2\left(4 * \dfrac{21}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11} + 1\right) \\[5ex] 2\left(\dfrac{84}{11}\ + \dfrac{11}{11}\right) \\[5ex] 2\left(\dfrac{84 + 11}{11}\right) \\[5ex] 2\left(\dfrac{95}{11}\right) \\[5ex] \dfrac{190}{11} $ | $ \underline{RHS} \\[3ex] 8 - 3(x - 5) \\[3ex] x = \dfrac{21}{11} \\[5ex] 8 - 3\left(\dfrac{21}{11} - 5\right) \\[5ex] 8 - 3\left(\dfrac{21}{11} - \dfrac{55}{11}\right) \\[5ex] 8 - 3\left(\dfrac{21 - 55}{11}\right) \\[5ex] 8 - 3\left(-\dfrac{34}{11}\right) \\[5ex] 8 - -\dfrac{102}{11} \\[5ex] 8 + \dfrac{102}{11} \\[5ex] \dfrac{88}{11} + \dfrac{102}{11} \\[5ex] \dfrac{88 + 102}{11} \\[5ex] \dfrac{190}{11} $ |

(26.)

$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $__Check__

$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $

$ \underline{LHS} \\[3ex] 4 - m \\[3ex] m = -4 \\[3ex] 4 - (-4) \\[3ex] 4 + 4 \\[3ex] 8 $ | $ \underline{RHS} \\[3ex] 2m + 16 \\[3ex] m = -4 \\[3ex] 2(-4) + 16 \\[3ex] -8 + 16 \\[3ex] 8 $ |

(27.)

$ 4(3p + 2) - 7 = 3(p - 2) \\[3ex] 12p + 8 - 7 = 3p - 6 \\[3ex] 12p - 3p = -6 - 8 + 7 \\[3ex] 9p = -7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] $__Check__

$ 4(3p + 2) - 7 = 3(p - 2) \\[3ex] 12p + 8 - 7 = 3p - 6 \\[3ex] 12p - 3p = -6 - 8 + 7 \\[3ex] 9p = -7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] $

$ \underline{LHS} \\[3ex] 4(3p + 2) - 7 \\[3ex] p = -\dfrac{7}{9} \\[5ex] 4\left(3 * -\dfrac{7}{9} + 2\right) - 7 \\[5ex] 4\left(-\dfrac{21}{9} + \dfrac{18}{2}\right) - 7 \\[5ex] 4\left(\dfrac{-21 + 18}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{3}{9}\right) - 7 \\[5ex] 4\left(-\dfrac{1}{3}\right) - 7 \\[5ex] -\dfrac{4}{3} - 7 \\[5ex] -\dfrac{4}{3} - \dfrac{21}{3} \\[5ex] \dfrac{-4 - 21}{3} \\[5ex] -\dfrac{25}{3} $ | $ \underline{RHS} \\[3ex] 3(p - 2) \\[3ex] p = -\dfrac{7}{9} \\[5ex] 3\left(-\dfrac{7}{9} - 2\right) \\[5ex] 3\left(-\dfrac{7}{9} - \dfrac{18}{9}\right) \\[5ex] 3\left(\dfrac{-7 - 18}{9}\right) \\[5ex] 3\left(-\dfrac{25}{9}\right) \\[5ex] -\dfrac{25}{3} $ |

(28.)

$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $__Check__

$ 4 - m = 2m + 16 \\[3ex] 4 - 16 = 2m + m \\[3ex] -12 = 3m \\[3ex] 3m = -12 \\[3ex] m = -\dfrac{12}{3} \\[5ex] m = -4 \\[3ex] $

$ \underline{LHS} \\[3ex] 4 - m \\[3ex] m = -4 \\[3ex] 4 - (-4) \\[3ex] 4 + 4 \\[3ex] 8 $ | $ \underline{RHS} \\[3ex] 2m + 16 \\[3ex] m = -4 \\[3ex] 2(-4) + 16 \\[3ex] -8 + 16 \\[3ex] 8 $ |