Ask question

Ask question

All categories

3 years ago

13

What is the upper quartile of the box plot graph?

1 answer:

julia-pushkina [17]3 years ago

3 0

Answer: Choice A) 90

The upper quartile, aka the third quartile (Q3), is visually represented by the right-side edge of the box, which is at 90 on the number line.

You might be interested in

Arrange the systems of equations that have a single solution in increasing order of the x-values in their solutions. 2x + y = 10

Sladkaya [172]

Answer: 4,0,2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

I need help ASAP!!!!!

Alla [95]

If we know that two sets of corresponding angles and the corresponding included sides are congruent in two triangles, what can we say about the triangles?

3 0

3 years ago

Which of the following statements must be true in order for the line represented by the equation y = mx + b to have a negative x

nekit [7.7K]

<span>The signs of the values of m and b<span> are the same.</span></span>

6 0

3 years ago

IRISSAK [1]

Your answer is A 1/15

6 0

4 years ago

Read 2 more answers

Total blood cholesterol level was measured for each of

laiz [17]

<h3>Answer: Mean = 218.9.</h3><h3>Median = 229</h3><h3>Mode = Zero mode.</h3>

Step-by-step explanation:

Given blood cholesterol level was measured for each of 8 adults (in mg/dL) are:

264, 191, 160, 148, 262, 212, 268, 246

In order to find the mean, we need to add all those 8 numbers and divide by 8.

Therefore, mean = $\frac{264+191+160+148+262+212+268+246}{8} =\frac{1751}{8} =218.9.$

<h3>Mean = 218.9.</h3>

In order to find the median, we need to arrange them in ascending order:

148, 160, 191, 212, 246, 262, 264, 268.

The middle most two values are 212 and 246.

Therefore, median = $\frac{212+246}{2} =\frac{458}{2} = 229.$

<h3>Median = 229.</h3>

$\mathrm{The\:mode\:is\:the\:term\:in\:the\:data\:set\:that\:appears\:the\:most.}$

$\mathrm{If\:there\:is\:more\:than\:one\:term\:that\:appears\:the\:most,\:then\:there\:is\:no\:mode.}$

$\mathrm{Count\:the\:number\:of\:times\:each\:element\:appears\:in\:the\:list}$

$\begin{pmatrix}148&160&191&212&246&262&264&268\\ 1&1&1&1&1&1&1&1\end{pmatrix}$

$\mathrm{The\:most\:common\:element\:is\:not\:unique,\:so\:there\:is\:no\:mode}$

<h3>Therefore, Mode = Zero mode.</h3>

5 0

3 years ago