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3 years ago

9

Lana put an end table with a triangular top next to her sofa. The base of the top is 26 in., and the height is 16 in.

2 answers:

Tresset [83]3 years ago

4 0

26in*16in/.5=208

208in2

asambeis [7]3 years ago

4 0

Answer:

Area of the triangular top is 208 in²

Step-by-step explanation:

We have, the triangular top has dimensions,

Base = 26 inches

Height = 16 inches.

As, area of a triangle = $\frac{1}{2}\times Base\times Height$

So, Area of the top = $\frac{1}{2}\times 26\times 16$

i.e. Area of the top = $\frac{416}{2}$

i.e. Area of the top = 208 in²

Hence, the area of the triangular top is 208 in².

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What is the approximate volume of a cone with a radius of 15 cm and a height of 4 cm? Round your answer to the nearest hundredth

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Answer:

Volume of a cone $=942.86cm^3$

Step-by-step explanation:

Given that the radius of a cone is 15cm and its height is 4cm

That is r=15cm and h=4cm

To find the volume of a cone :

volume of a cone $=\frac{\pi r^2h}{3}$ cubic units

Now substitute the values in the formula we get

volume of a cone $=\frac{(\frac{22}{7}) (15)^2(4)}{3}$

$=\frac{(\frac{22}{7}) (225)(4)}{3}$

$=\frac{19800}{21}$

$=942.857$

Now round to nearest hundredth

$=942.86$

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Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

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Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

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