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3 years ago

Which shows the measures in ascending order for the data below?

Mathematics

2 answers:

MrMuchimi3 years ago

3 0

Answer:

Population size: 9

Median: 14

Minimum: 11

Maximum: 18

First quartile: 11

Third quartile: 16.5

Interquartile Range: 5.5

Outliers: none

-BARSIC- [3]3 years ago

3 0

Answer:

Step-by-step explanation:

1 2

Range

18-11=7

Mode

Mean

13.9

Median

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Which numbers are irrational?

Phoenix [80]

V15 and 0.151155111555...

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3 years ago

How many real solutions does p^2-5p+7=0 have

jekas [21]

Let's work out the discriminant b^2 - 4ac:-

D = (-5)^2 - 4*1*7 = 25 - 28 = -3

A negative discriminant shows that there are NO real solutions Answer

8 0

3 years ago

What is √101.6064 in to 1 decimal place and in standard form?<br>NEED ASAP

Dimas [21]

Note that

10² = 100

0.08² = 0.0064

and

101.6064 = 100 + 1.6 + 0.0064

It also happens that

1.6 = 2 • 10 • 0.08

which means

101.6064 = 100 + 1.6 + 0.0064

101.6064 = 10² + 2 • 10 • 0.08 + 0.08²

101.6064 = (10 + 0.08)²

101.6064 = 10.08²

which means

√101.6064 = 10.08

which, to one decimal place, is approximately 10.1.

3 0

3 years ago

What is the common difference in the following arithmetic sequence?

Ivan

The answer is D

-12+11=-1

-1+11=10

10+11=21

You add because it is an arithmatic sequence rather than a geometric sequence

7 0

3 years ago

Read 2 more answers

How do I find the Velocity and How long will it take for the ball to reach it's maximum height?

Colt1911 [192]

So hmm check the picture below

$\bf \qquad \textit{initial velocity}\\\\
h = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\ 
\begin{cases}
v_o=\textit{initial velocity of the object}\to &64\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds} \end{cases}\\\\
-----------------------------\\\\$

$\bf \textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+64t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

part 1)

it takes $\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds$

part 2)

$\bf \textit{now, doubling }v_o\\\\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &128\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds}\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+128t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

it will reach the maximum height at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet$

how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$ ?

subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became

4 0

3 years ago