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Hatshy [7]
2 years ago
13

Can someone please help me ASAP

Mathematics
1 answer:
fiasKO [112]2 years ago
3 0

Answer:

Step-by-step explanation:

AB = 5

AC = 6

BC = √(5² + 6²) = √61 ≈ 7.8 units

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Answer:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

Step-by-step explanation:

Given

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}

Required

Fill in the box

From the question, the range is:

Range = 60

Range is calculated as:

Range =  Highest - Least

From the box, we have:

Least = 1

So:

60 = Highest  - 1

Highest = 60 +1

Highest = 61

The box, becomes:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question:

IQR = 20 --- interquartile range

This is calculated as:

IQR = Q_3 - Q_1

Q_3 is the median of the upper half while Q_1 is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}

<u>Upper half</u>

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

Median = \frac{N + 1}{2}th

Where N = 7

So, we have:

Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th

So,

Q_3 = 4th item of the upper halves

Q_1= 4th item of the lower halves

From the upper halves

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

<u></u>

We have:

Q_3 = 32

Q_1 can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

IQR = Q_3 - Q_1

Where Q_3 = 32 and IQR = 20

So:

20 = 32 - Q_1

Q_1 = 32 - 20

Q_1 = 12

So, the lower half becomes:

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}

From this, the updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question, the median is:

Median = 22 and N = 14

To calculate the median, we make use of:

Median = \frac{N + 1}{2}th

Median = \frac{14 + 1}{2}th

Median = \frac{15}{2}th

Median = 7.5th

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

Mode = 18

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

7th = 18

The 8th item is calculated as thus:

Median = \frac{1}{2}(7th + 8th)

22= \frac{1}{2}(18 + 8th)

Multiply through by 2

44 = 18 + 8th

8th = 44 - 18

8th = 26

The updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question.

Mean = 26

Mean is calculated as:

Mean = \frac{\sum x}{n}

So, we have:

26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}

Collect like terms

26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}

26= \frac{ 2nd + 12th+304}{14}

Multiply through by 14

14 * 26= 2nd + 12th+304

364= 2nd + 12th+304

This gives:

2nd + 12th = 364 - 304

2nd + 12th = 60

From the updated box,

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

We know that:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

2nd = 3

12th = 57

i.e.

2nd + 12th = 60

3 + 57 = 60

So, the complete box is:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

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Answer:

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Step-by-step explanation:

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