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3 years ago

11.7÷(0.7+0.6)×2 Please help

Mathematics

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

0.7+0.6 = 1.3

1.3x2 = 2.6

11.7÷2.6 = 4.5

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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)

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

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

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:

Lower half:

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

From this, the updated values of the box is:

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