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Mama L [17]
11 months ago
6

Suppose we have a bottomless bag of 3 different kinds of balls: blue, green, and red. How many possibilities are there to choose

3 balls and arrange them in a line.
Mathematics
1 answer:
sleet_krkn [62]11 months ago
6 0

The number of possibilities that are there to choose 3 balls is 6.

<h3>How to illustrate the information?</h3>

From the information, we have a bottomless bag of 3 different kinds of balls: blue, green, and red.

The number of possibilities that are there to choose 3 balls will be:

= 3!

= 3 × 2 × 1

= 6 possibilities.

Therefore, the number of possibilities that are there to choose 3 balls is 6.

Learn more about permutations on:

brainly.com/question/4658834

#SPJ1

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Goshia [24]
You should write numbers in as many ways as you possibly can to make new connections in your brain. Knowing how to write numbers in many different ways can help you solve complex problems more easily. Doing this can also reinforce the mathematical principles and logic you have memorised.

Writing one in many different ways:

1=1/1=2/2=3/3=4/4=(-1)/(-1)=(-2)/(-2)

=1.0=1.00=1.000=(1/2)+(1/2)=(1/3)+(1/3)+(1/3)

=(1/4)+(1/4)+(1/4)+(1/4)

Writing a half in many different ways:

1/2=(1/4)+(1/4)=(1/6)+(1/6)+(1/6)

=(1/8)+(1/8)+(1/8)+(1/8)=4*(1/8)

=2/4=3/6=4/8=5/10=0.5=0.50

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6 0
2 years ago
I need help please help me thanks please
Vitek1552 [10]

so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})

\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{4}(x-\stackrel{x_1}{1})\\\\\\y-3=4x-4\implies y = 4x-1\implies \blacktriangleright  y = 4x+(-1)\blacktriangleleft

4 0
2 years ago
A circle has a radius of 2.5 centimeters and a central angle AOB that measures 90°. What is the area of sector AOB? Use 3.14 for
cupoosta [38]

The area of any circle is        (pi) x (radius)²  .

The area of the WHOLE circle in the question is

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The full circle is made by 360° of central angles.

90° of central angle makes a pie-wedge (sector)
that's exactly one fourth of the circle, so its area
is one fourth of the area of the whole circle.

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Rounded to the nearest tenth:   4.9 cm² .

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MatroZZZ [7]

Answer:

153 In.

Step-by-step explanation:

Add the area area of each of the individual shapes together

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7 0
3 years ago
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