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3 years ago

DISTANCE FORMULA PRACTICE 1-3

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

Answer:

1. 12 units

2. 8.9 units

3. 10

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steps

1.√ ( − 3 − 9 ) ^2 + ( 5 − 5 ) ^2

√ ( − 12 ) ^2 + 0

√ 144

√ 144 ≈ 12

************************************************

2. √ ( 7 − (− 1) ) ^2 + ( 10 − 6 ) ^2

√ ( 7 + 1 )^ 2 + ( 10 − 6 ) ^2

√ ( 8 )^ 2 + ( 4 ) ^2

√ 64 + 16

√ 80 ≈ 8.94427190999916

round up to 8.9

*****************************************************

3. √ ( 6 − 0 ) ^2 + ( 0 − 8 ) ^2

√ ( 6 ) ^2 + ( − 8 ) ^2

√ 36 + 64

√ 100 ≈ 10

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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:

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

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From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

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$Least = 1$

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$60 = Highest - 1$

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The box, becomes:

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

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This is calculated as:

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$Q_3$ 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)

Lower half:

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

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

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

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We have:

$Q_3 = 32$

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So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

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So, the lower half becomes:

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

From this, the updated values of the box is:

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

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

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$Median = \frac{15}{2}th$

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

The 7th and 8th items are blanks.

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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:

$7th = 18$

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$Mean = 26$

Mean is calculated as:

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

So, we have:

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

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

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

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This gives:

$2nd + 12th = 364 - 304$

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