All categories

2 years ago

I can’t seem to find this, I’m having trouble with this question! It would be nice if someone could help! Thanks

Mathematics

1 answer:

Elenna [48]2 years ago

8 0

Answer:

-456 I think but I could be wrong though

You might be interested in

R/3 + 5 < 8<br><br> please help

Feliz [49]

Answer:

r<9

Step-by-step explanation:

r/3 + 5 < 8

Subtract 5 from each side

r/3 + 5-5 < 8-5

r/3 < 3

Multiply each side by 3

r/3 *3 <3*3

r<9

3 0

3 years ago

What is the value of the function f(x) = 2x – 5 when x = 22? A17 B49 C105 D39

enyata [817]

<span>f(x) = 2x – 5

</span><span>f(22) = 2(22) – 5 = 44 -5 = 39

answer

</span><span>D39</span>

8 0

3 years ago

Three runners are training for a marathon. One day, they all run about ten miles, each at their own constant speed. Which of the

kotykmax [81]

Answer:

Runner 1 has the fastest pace

Step-by-step explanation:

The first one we need to find the slope

m = (y2-y1)/(x2-x1) = (40-0)/(4-0) =40/4 = 10 minutes per mile

The second is 8.5 minutes per mile

The third we need to find the slope

m = (y2-y1)/(x2-x1) = (90-72)/(10-8) =18/2 = 9 minutes per mile

6 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