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## MTEL General Curriculum Mathematics Practice

Question 1 |

#### The picture below represents a board with pegs on it, where the closest distance between two pegs is 1 cm. What is the area of the pentagon shown?

Question 2 |

#### Four children randomly line up, single file. What is the probability that they are in height order, with the shortest child in front? All of the children are different heights.

\( \large \dfrac{1}{4}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |

\( \large \dfrac{1}{256}
\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |

\( \large \dfrac{1}{16}\) Hint: Try a simpler question with 3 children -- call them big, medium, and small -- and list all the ways they could line up. Then see how to extend your logic to the problem with 4 children. | |

\( \large \dfrac{1}{24}\) Hint: The number of ways for the children to line up is \(4!=4 \times 3 \times 2 \times 1 =24\) -- there are 4 choices for who is first in line, then 3 for who is second, etc. Only one of these lines has the children in the order specified. |

Question 3 |

#### A family has four children. What is the probability that two children are girls and two are boys? Assume the the probability of having a boy (or a girl) is 50%.

\( \large \dfrac{1}{2}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |

\( \large \dfrac{1}{4}\) Hint: How many different configurations are there from oldest to youngest, e.g. BGGG? How many of them have 2 boys and 2 girls? | |

\( \large \dfrac{1}{5}\) Hint: Some configurations are more probable than others -- i.e. it's more likely to have two boys and two girls than all boys. Be sure you are weighting properly. | |

\( \large \dfrac{3}{8}\) Hint: There are two possibilities for each child, so there are \(2 \times 2 \times 2 \times 2 =16\) different configurations, e.g. from oldest to youngest BBBG, BGGB, GBBB, etc. Of these configurations, there are 6 with two boys and two girls (this is the combination \(_{4}C_{2}\) or "4 choose 2"): BBGG, BGBG, BGGB, GGBB, GBGB, and GBBG. Thus the probability is 6/16=3/8. |

Question 4 |

#### The histogram below shows the number of pairs of footware owned by a group of college students.

#### Which of the following statements can be inferred from the graph above?

## The median number of pairs of footware owned is between 50 and 60 pairs.Hint: The same number of data points are less than the median as are greater than the median -- but on this histogram, clearly more than half the students own less than 50 pairs of shoes, so the median is less than 50. | |

## The mode of the number of pairs of footware owned is 20.Hint: The mode is the most common number of pairs of footwear owned. We can't tell it from this histogram because each bar represents 10 different numbers-- perhaps 8 students each own each number from 10 to 19, but 40 students own exactly 6 pairs of shoes.... or perhaps not.... | |

## The mean number of pairs of footware owned is less than the median number of pairs of footware owned.Hint: This is a right skewed distribution, and so the mean is bigger than the median -- the few large values on the right pull up the mean, but have little effect on the median. | |

## The median number of pairs of footware owned is between 10 and 20.Hint: There are approximately 230 students represented in this survey, and the 41st through 120th lowest values are between 10 and 20 -- thus the middle value is in that range. |

Question 5 |

#### Use the samples of a student's work below to answer the question that follows:

#### This student divides fractions by first finding a common denominator, then dividing the numerators.

\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\)#### Which of the following best describes the mathematical validity of the algorithm the student is using?

## It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.Hint: Don't be so rigid! Usually there's more than one way to do something in math. | |

## It got the right answer in these three cases, but it isn‘t valid for all rational numbers.Hint: Did you try some other examples? What makes you say it's not valid? | |

## It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.Hint: Lowest terms doesn't affect this problem at all. | |

## It is valid for all rational numbers, as long as the divisor is not zero.Hint: When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts. |

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