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2 years ago

13

Helppp quickkkkkkk enough points

2 answers:

xenn [34]2 years ago

8 0

Answer:

A or 57 7/9ft3

Step-by-step explanation:

13/3x10/3x4/1 = 520/9=57 7/9

Fynjy0 [20]2 years ago

7 0

Answer:

$a=2\\\\ b = 20\\\\ c = 47$

Step by step explanation:

$a(x) = 2(x+5)^2 -3\\\\~~~~~~~=2(x^2 +2 \cdot 5 \cdot x +5^2) -3\\\\~~~~~~~=2(x^2 +10x+25) -3\\\\~~~~~~~=2x^2 +20x +50 -3 \\\\~~~~~~~=2x^2 +20x +47\\\\\text{By comparing}~ 2x^2 +20x +47~ \text{with the standard form}~ ax^2 +bx+c,\\\\a=2,~ b = 20,~ c = 47$

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Find the area of the shaded region. Use 3.14 for pi. Round to the nearest tenth.<br> r = 15<br> 110

Svetradugi [14.3K]

Answer:

110/360=11/36 (I will get back to this)

The area of the circle would be 225 pi using the formula pi r^2. So you multiply 225pi by 11/36 to find the area. This rounds to 215.98

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3 years ago

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Can a triangle have 2 right angles

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No they can only have one right angle

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123 x 10 to the -6 power

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The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

<u>For Second Half:</u>

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

<u>Interquartile Range:</u>

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago

Henry steps onto an elevator at the first floor. He goes up to the 9th floor and then comes down 2 floors. He then goes up 4 flo

Anika [276]

Answer:

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3 years ago

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