You would time the denominator by 3 so 11*3 would be 33 then times the numerator by the same so 6*3 soot would be 18/33
Well, there is nothing following, but if there is anything about K equal to or greater 4, or K being>2 than thats it.
Answer:
V = 300x + -70x2 + 4x3
Step-by-step explanation:
Simplifying
V = (20 + -2x)(15 + -2x)(x)
Reorder the terms for easier multiplication:
V = x(20 + -2x)(15 + -2x)
Multiply (20 + -2x) * (15 + -2x)
V = x(20(15 + -2x) + -2x * (15 + -2x))
V = x((15 * 20 + -2x * 20) + -2x * (15 + -2x))
V = x((300 + -40x) + -2x * (15 + -2x))
V = x(300 + -40x + (15 * -2x + -2x * -2x))
V = x(300 + -40x + (-30x + 4x2))
Combine like terms: -40x + -30x = -70x
V = x(300 + -70x + 4x2)
V = (300 * x + -70x * x + 4x2 * x)
V = (300x + -70x2 + 4x3)
Solving
V = 300x + -70x2 + 4x3
Solving for variable 'V'.
Move all terms containing V to the left, all other terms to the right.
Simplifying
V = 300x + -70x2 + 4x3
The outlier of a dataset is a data element that is relatively far from the remaining data elements
- <em>99 is an outlier of pet group</em>
- <em>See attachment for the parallel box plots</em>
<u>(a) Prove that 99 is an outlier for Pet</u>
We have:
<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>
The quartiles positions are:
So, we have:
From the pet group:
The data elements at the 4th and 12th positions are 68 and 79
So, we have:
The lower and upper limits of the outlier are:
So, we have:
This means that data below 51.5 or above 95.5 are outliers.
<em>Hence, 99 is an outlier because 99 is greater than 95.5</em>
<u>(b) The parallel box plot</u>
The three groups are:
<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>
<em>Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82 </em>
<em>Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99</em>
<em />
See attachment for the parallel box plots
Read more about box plots and outliers at:
brainly.com/question/14940764
Answer:
4x - 8 + 4y
Step-by-step explanation:
4(x - 2 + y)
4x - 8 + 4y
You multiply each term in the parenthesis by the number in front of it.