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makkiz [27]
3 years ago
5

Which statement is supported by the histograms? The two data sets have the same distribution and same range. The two data sets h

ave the same distribution but different ranges. The two data sets have different distributions but the same range. The two data sets have different distributions and different ranges.
Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:

The two data sets have different distributions and different ranges.

Step-by-step explanation:

When you subtract for both of the ranges they are different numbers. So that's the second part. The numbers along the bottom are different too so that is the different distributions.

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Evgesh-ka [11]
Hi, the mean for the numbers is 29, B =D
4 0
3 years ago
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There are some beads in the box. 40% of them were blue and the rest were red. Adam used all the blue beads and 25% of the red be
Shalnov [3]

Answer:

65% of the beads were used to make the bracelet.

Step-by-step explanation:

40% of the beads were blue which means 60% of the beads were red. If all the blue beads were used that means 40% was used so you add the 25% of the red beads that were used which is why the answer is 65%.

7 0
3 years ago
Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
Hitman42 [59]
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

7 0
3 years ago
Need help with numbers 1,2&3 pls asap
sineoko [7]
-4(-x-10)= 4x• 40
11g+4-6g+7s-10 =16g+6-7s
3(10+9x)+11-3x= 13x+19
5 0
3 years ago
Solve and graph: 7x-6&gt;22<br>(pls explain your answer, im confused on this)​
grandymaker [24]

Answer:

x>4

Step-by-step explanation:

Solving an inequality is very similar to solving an equation:

Given: 7x-6>22

Add 6 to both sides: 7x>28

Divide both side by 7: x>4

x>4 means that all x-values that are greater than 4 will work. Therefore you would write an open dot at x=4 on the number line to show that 4 isn't included, and draw an arrow going right to show all possible solutions greater than 4.

6 0
3 years ago
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