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3 years ago

Which of the following is not needed when making a box plot?

Mathematics

2 answers:

algol133 years ago

3 0

The mean is not needed to make a box plot.

I hoped this helped!

Fofino [41]3 years ago

3 0

Answer: A) Mean

You use the median as the center of the boxplot. The first and third quartiles (Q1 and Q3) make up the left and right edges of the box. The min and max are the furthest points on the left and right side of each whisker. The mean is not used at all. With outliers, the mean is skewed. If you have any outliers then its best to use the median instead.

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Please help meee with my math i am so confused!!! Thank youuuu

Rudik [331]

Answer:

Step-by-step explanation:

x decreases w/o bound, so x becomes a very large -ve no.

f(x)=5^(x-1) becomes 5^(large -ve no.) which will be almost zero

x increases w/o bound, so x becomes a very large +ve no.

f(x)=5^(x-1) becomes 5^(large +ve no.) which will keep increasing w/o bound

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3 years ago

Read 2 more answers

Half note as a decimal?

GrogVix [38]

0.5 or .5 (remember, half a whole note is a half note)

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3 years ago

What is the area of this rectangle?

nataly862011 [7]

Answer: 18 units squared

Step-by-step explanation:

To find area of a rectangle use A=lw

we can count the units for both and get...

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So 18 units squared

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2 years ago

Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

miss Akunina [59]

Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

Equation:

$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

$y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

7 0

3 years ago

Read 2 more answers

Hippalectryon

horrorfan [7]

Since the stadium is 6 mi from the practice field which is itself 1 mi from the house, it can either be (1+6)=7 mi North or (1-6)=-5 -> 5 mi South depending on the North/South direction.

the answer is thus A

7 0

3 years ago