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AleksandrR [38]

3 years ago

7

NEED HELP ASAP!!! I'LL GIVE BRAINLIEST FOR THE CORRECT ANSWER!!!

1 answer:

Scorpion4ik [409]3 years ago

3 0

Answer: 8957545658374365

Step-by-step explanation:

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Which of the answer choices could be the one missing data item for the following set, if the mode is 12?

crimeas [40]

Answer:

D

Step-by-step explanation:

first of all, what is mode: it is the highest occuring number in a set of numbers.

now since 12 is supposed to be the mean and it occurs twice but 9 also occurs twice so for 12 to be the mode, we must add one more 12.

therefore the answer is 12

8 0

3 years ago

Read 2 more answers

T-activity-player/star/start

wolverine [178]

Answer:

$2400

Step-by-step explanation:

as you realize, it is going down by $400 per year

6000, 5600, 5200, 4800, 4400, 4000, 3600, 3200, 2800, 2400
1. 2. 3. 4. 5. 6. 7. 8. 9. 10

it would be $2400

3 0

2 years ago

What is the aquare root of 200

Anton [14]

Answer:

$=10\sqrt{2}$

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors.

Like this :

5 0

3 years ago

A thin wire has the shape of the first-quadrant part of the circle with center the origin and radius q. If the density function

irina [24]

Answer:

Step-by-step explanation:

Check attachment for solution

5 0

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

6 0

2 years ago