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hjlf
3 years ago
9

There are 5 yellow buttons, 2 green buttons, and 3 red buttons in a bag. Find the following probabilities A. Find the probabilit

y of pulling yellow or red button. Answer
B. find the probability of pulling a green button, then a red button with replacement. Answer and C. Find the probability of pulling a red button, then a yellow button WITHOUT replacement. Answer
Mathematics
2 answers:
goldenfox [79]3 years ago
8 0
80%
6%
16.7%
answerrsss
ludmilkaskok [199]3 years ago
5 0
There are 10 buttons total then pulling a green button out of the total is 2/10 and a red is 3/10 and pulling a red button is 3/10 and a yellow button 5/9
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if you draw one card from a standard deck, what is the probability of drawing a diamond or a red card?​
Aleks [24]

Answer:

3/4 probability

probability to draw a red card = 11/26

probability to draw a diamond =1/4

probability of drawing one or other card = 35/52

# actually i am not sure

4 0
3 years ago
What is 1 1/3 times 1 4/9 in fraction form
nikitadnepr [17]

Answer:

52 / 27

Step-by-step explanation:

3 0
3 years ago
Solve x 2 + 9x + 8 = 0 by completing the square. What are the solutions?
vivado [14]

x^2+9x+8=0

 (x+1)(x+8)=0

x+1=0

x=-1

x+8 = 0

x=-8


 solutions are -1 and -8

3 0
3 years ago
Three people go out to lunch. They decide
kykrilka [37]

Answer:

$23.00 each

Step-by-step explanation:

$69 / 3 people = $23 from each person

7 0
2 years ago
Can anyone figure this out?
Verizon [17]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

5 0
3 years ago
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