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3 years ago

Circle has a circumference of 59 cm using 3.14 as an approximation for π (pie)

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

6 0

I believe the radius of the circle would be about 18.79 (rounded).
Hope this helps ;}

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9. A box of cereal is packaged and sold at a local

emmasim [6.3K]

Answer:

288 Cubic Inches

Step-by-step explanation:

Assuming the cereal takes up 80% of the total volume of the box, you would take 80% of the total volume.

First, do Base x Width x Height for the volume of the box.

10 inches x 3 inches x 1 foot (1 foot is 12 inches)

10 x 3 x 12 = 360

Now, take 80% of 360 cubic inches.

80% is 0.8 times a number

0.8 x 360 = 288

288 Cubic Inches

8 0

2 years ago

I'm doing simplifying expressions with exponents

VladimirAG [237]

Answer:

Expressing a number in a power of 2 means that you have to figure out how many times you should multiply 2 by itself so the product will be that number.

The number is expressed as 2^n

For example:

Express 16 as a power of 2:

To get 16, you have to multiply 2 by 4 times.

2 × 2 × 2 × 2 = 16

16 = 2^4

Then the answer is 2^4

7 0

3 years ago

Consider the function below.

Brut [27]

Answer:

Step-by-step explanation:

6 0

2 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