All categories

3 years ago

Here is the next question i need help with PLEASE HELP

Mathematics

2 answers:

Rasek [7]3 years ago

8 0

Rc = 60×2=120
you chose the right answer

pogonyaev3 years ago

7 0

Each angle is 60. so doubl;e that 60 * 2 = 120

You might be interested in

Simplify the expression. <br> −9(−2 − 8) + (−16) −8

Xelga [282]

Answer:

Step-by-step explanation:

Open the parentheses, 18 + 72 - 16 - 8 = 66

7 0

3 years ago

Read 2 more answers

6y+12=-2x <br> x+3y=-6 <br> In y form

Vlada [557]

Answer:

both equations reduce to y = -1/3x -2

Step-by-step explanation:

In each case, subtract the term not containing y and divide by the coefficient of y.

1. 6y = -2x -12 . . . . 12 was subtracted from both sides

y = -1/3x -2 . . . . . . the equation was divided by 6

2. 3y = -x -6 . . . . . . x was subtracted from both sides

y = -1/3x -2 . . . . . the equation was divided by 3

4 0

3 years ago

A. The number 86,362,575 rounded to the nearest hundred is

Serjik [45]

Answer:

86362600

Step-by-step explanation:

We want to round to the hundreds place

We look at the 7 in the tens place. Since it is 5 or greater we round up

86,362,575 round to 86362600

7 0

3 years ago

Read 2 more answers

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