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1 year ago

Find the area of the Shaded region below in all of the smaller squares are identical in size

Mathematics

1 answer:

Lelechka [254]1 year ago

7 0

Area of the shaded region = 80 in²

Explanation:

The outer diagram is a square

Area of the square = length × length

length = 12in

Area of the square = 12in × 12in

Area of the square = 144in²

There are 3 smaller squares in each row and column:

The length in each row and colum = 12

the length o fone small box = 12/3 = 4

Area of the smaller box = 4in × 4in = 16 in²

There are 4 boxes unshaded

Area of 4 unshaded boxes = 4 × 16 in² = 64in²

Area of the shaded region = Area of the outer box - area of the unshaded boxes

Area of the shaded region = 144 - 64

Area of the shaded region = 80 in²

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Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

$321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320$

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}$

STEP 3: Calculate the mean

$\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}$

STEP 4: Calculate the Standard deviation

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}$

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

$\begin{gathered} \text{IQR}=Q_3-Q_1 \\ \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\ \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}$

STEP 6: Find the Interquartile Range