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nikklg [1K]
3 years ago
11

Melissa has a collection of different colored hair bows. Some of the hair bows are big, and some of the hair bows are small. The

chart below shows how many of each type of hair bow is in Melissa's collection. What is the probability that the hair bow selected will be big and pink?

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
4 0

Answer:

7/55 or 12.727272727273%

Step-by-step explanation:

7/55 or 12.727272727273%

7 big and pink

so 7/?

?= amount of bows

and them up and you get 55

7/55

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7 more than the quantity 3 times 2
Zolol [24]

Answer:

13

Step-by-step explanation:

(3 times 2)+7=x

6+7=x

x=13

5 0
3 years ago
One third of the result of three times a number that is increased by 12
VikaD [51]

Answer:

Step-by-step explanation:

1/3 (3x+12)

1/3×3x+1/3×12

x+4

6 0
2 years ago
Question is in the picture
Thepotemich [5.8K]

Answer:

$9.00

Step-by-step explanation:

If the price is at 30% discount, this means it is 100-30 = 70% of the original price.

70 \%  \ of \ original \ price = \$ 6.30\\\\\ original \ price = 6.30 \div 70 \%\\\\= 6.30 \times \frac{100}{70}\\\\=\frac{630}{70}\\\\= 9

Hence, original price is $9.00.

3 0
3 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
X<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B3%7D%20%20-%2012%20%3D%20%20-%202" id="TexFormula1" title=" \frac{x}{
gulaghasi [49]
Wish fb ee snsjsjwbw e Bush’s use Brisbane Meeks
6 0
3 years ago
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