All categories

3 years ago

15 POINTS NO CAP PLEASE HELP ME NEED RIGHT ANSWER

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

5 0

Answer:

y = -x + 3

Step-by-step explanation:

The line is going down so the slope will be NEGATIVE

The line starts at (0, 3) which means the y-intercept is 3

The slope is -1 or -x

dusya [7]3 years ago

3 0

Answer: c

Step-by-step explanation:

You might be interested in

jenny peddled her bike for 11 miles in 45 minutes if she traveled the same speed how far would she reach in 2 hours?

professor190 [17]

60 * 2 = 120
120 / 45 = 2.6...7
2.6 * 45 = 29.3... (3 repeating) mi.

Messed that up!

11 / 45 = .24...
.24 * 120 = 28.8<u />

8 0

3 years ago

Read 2 more answers

00013 in scientific notation

Svet_ta [14]

1.3 x 10^1

hope that helped
let me know if i am right please

7 0

3 years ago

Trick or treat?<br><br> Pls answer ;-;

BARSIC [14]

Treat??

hope this helps lol

3 0

3 years ago

Read 2 more answers

Which is equivalent to 8X- Y + 3X

Zarrin [17]

Answer:

11x-y or 5x+y.

Step-by-step explanation: I hope this helps, I really don't know what the question means, since it not that specific.

6 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