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mr Goodwill [35]

2 years ago

14

Cindy has a poster in the shape of a parallelogram. What is the area of this parallelogram? 2.85 ft² 3.60 ft² 4.35 ft² 15.21 ft²

Parallelogram A B C D with side A D parallel to side B C and side A B is parallel to side D C. Point E is on side A B and is connected to point D by a dotted line. Point F is on side DC and connected to B by a dotted line. Angle D E B and angle F B E are right angles. A E and F C are 0.5 feet each. E B is 1.9 feet. F B is 1.5 feet.

1 answer:

katovenus [111]2 years ago

4 0

Answer:

The area is 6 ft^2

Step-by-step explanation:

The base of the parallelogram is AE+EB, which is .5+1.9, and that equals 2.4.

The height of the parallelogram is FB, because it's a perpendicular line straight down (know that from the right angle formed), so it must be the height. That is 1.5.

The formula for area is base x height, which is 2.4x2.5, which equals 6.

Therefore, area of the parallelogram is 6 ft^2

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y = 3/4x - 8/3

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3 people work full time together on a project, but their total time on the project is to be equivalent to only 1 person working

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3 years ago

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3 years ago

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

tresset_1 [31]

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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