$\text{Let the side of square 1, square 2 and square 3 be x, y and z respectively.}\\\\\text{Given that,}\\\\\text{Perimeter of square 1,~~ p =100 units.}\\\\\text{So}~ x=\dfrac p 4 = \dfrac{100}4 =25~\text{units.}\\\\\\\text{Area of square 2, a = 225 units}^2\\\\\text{So}~y = \sqrt{a} = \sqrt {225} = 15~ \text{units.}\\\\\text{Use Pythagorean theorem to find z}^2\\\\x^2 =y^2 +z^2 \\\\\implies 25^2 = 15^2 +z^2\\\\\implies z^2 = 625 - 225 = 400\\\\\\$

$\text{Hence the area of square 3 is 400 units}^2.$

8 0

3 years ago

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A recipe requires 3 cups of flour to make 27 dinner rolls. How much flour is needed to make 9 rolls?

Rudik [331]

It takes 1 cup of flour.

8 0

3 years ago

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3 questions in 1

MrRissso [65]

Answer:

1. D. 20, 30, and 50

2. A. 86

3. B. 94

Step-by-step explanation:

1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.

The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.

Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.

Thus:

Q1 = (85 + 87)/2 = 86

Q3 = (93 + 95)/2 = 94

IQR = Q3 - Q1 = 94 - 86

IQR = 8

Outliers in the data set are data values below the lower limit or above the upper limit.

Let's find the lower and upper limit.

Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74

The data values below the lower limit (74) are 20, 30, and 50

Let's see if we have any data value above the upper limit.

Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106

No data value is above 106.

Therefore, the only outliers of the data set are:

D. 20, 30, and 50

2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.

Thus:

Q1 = (85 + 87)/2 = 86

3. Q3 = (93 + 95)/2 = 94

3 0

2 years ago