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3 years ago

If a = b and b = c, which statement must be true?

Mathematics

2 answers:

murzikaleks [220]3 years ago

7 0

Answer:

D.- ac

Step-by-step explanation:

mark me as brainliest pls

Anestetic [448]3 years ago

5 0

Some text *, On the photo answer to ur question

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Jarods house is 17 miles from the soccer field there are 5,280 how many feet from the soccer field is jarods house

klemol [59]

Answer: 89,760 miles

Step-by-step explanation: 5,280*17=89,760.

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3 years ago

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12345 [234]

Answer:

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3 years ago

Multiply. Write answer in box below.<br> (3x - 4)(x+4)

adell [148]

Answer:3x^2+8x-16

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2 years ago

Given a line segment with endpoints A(16, 8) and B(1, 3) what are the coordinates of the

blondinia [14]

Answer:

The line segment partitioned two-fifths from A to B is (10,6)

Step-by-step explanation:

First point from A to B is (16,8)

than find the difference between A to B i.e B - A

(1,3)-(16,8) = (-15,-5)

To measure the (2/5) difference we will multiply (-15,-5) with (2/5) which is equal to (-6,-2)

Now Add the difference to the first coordinate (point A) gives

Point of division = (16,8)+(-6,-2)

Point of division = (16-6, 8-2)

Point of division = (10,6)

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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