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2 years ago

Who play's Little big snake "https://littlebigsnake.com/"

Mathematics

1 answer:

daser333 [38]2 years ago

5 0

Answer:

I have never heard of it sorry

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The inequality x ≥ −10 represents the possible values for x on a number line.

Leokris [45]

Answer:

-11 and lower

Step-by-step explanation:

it's saying that -10 is as low as the number can possibly be or equal to it.

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3 years ago

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(Will give brainliest)

Gemiola [76]

C) The king expected the statue to crumble.
I believe

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3 years ago

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Can 0.2 0.6 3 and 9 make 24?

saveliy_v [14]

Answer:

Yes

Step-by-step explanation:

9 x 3 - (0.6/0.2) Simplify the parentheses

9 x 3 - 3 Multiply 9 by 3

27 - 3 Subtract

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3 years ago

1/2 and a fraction equivalent to 2/6

makkiz [27]

Answer:1/2 if you multiply 2 to the top number then multiply 3 with the bottom

Step-by-step explanation:

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3 years ago

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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:

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