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viva [34]
3 years ago
14

A bag contains 2 red marbles and 3 black marbles. If Abby picks a marble without looking, returns it to the bag, and then draws

a second marble, what is the probability that both marbles are red? Give the answer as a fraction in simplest form.
Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
8 0
Answer: 4/25

=========================================

Explanation:

Let's define two events A and B
A = event of selecting red on the first draw
B = event of selecting red on the second draw

P(A) is the notation that means "probability of event A occurring"

P(A) = 2/5 because there are 2 red marbles out of 5 total (2 red + 3 black = 5)
Similarly, P(B) = 2/5 as well because A and B deal with the same color red, and because Abby put the first marble back

Multiply the probabilities
P(A and B) = P(A)*P(B) ... see note below
P(A and B) = (2/5)*(2/5)
P(A and B) = (2*2)/(5*5)
P(A and B) = 4/25 which is the answer

Note: the equation used is only valid if events A and B are independent, which they are in this case. The fact we put the marble back means the chances of picking red are the same as before.

Illusion [34]3 years ago
6 0
The correct answer is 4/25 i just did this question
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Answer:

\frac{1}{990}

Step-by-step explanation:

<u>The full question:</u>

<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current​ positions?"</em>

<em />

<em />

The permutation of choosing 3 members from a group of 11 would be:

P(n,r) = \frac{n!}{(n-r)!}

Where n would be the total [in this case n is 11] & r would be 3

Which is:

P(11,3) = \frac{11!}{(11-3)!}=\frac{11!}{8!}=11*10*9=990

So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:

1/990

8 0
3 years ago
A ride on the roller coaster costs 4 tickets while the boat ride only costs 3 tickets. Michael went on the two rides a total of
True [87]
Let x = roller coaster ticket
let y = boat ride ticket

x + y = 10
4x + 3y = 37

x = 10 - y
4x + 3y = 37
4(10-y) + 3y = 37
40 - 4y + 3y = 37
-y = 37 - 40
-y = -3
y = -3/-1
y = 3

x = 10 - y
x = 10 - 3
x = 7

x = 7 ; y = 3

4x + 3y = 37
4(7) + 3(3) = 37
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37 = 37 
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Answer:

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Step-by-step explanation:

DIFFERENTIATE W.R.T. B is a different method entirely

We simply add together the numerators and set with 2cos

then keep this number and add to sinb and square it.

then repeat initial 2 + cosb ^2 but instead of multiplying its add.

Then set the whole division to -sin (2b) squared then +1

<em> −  2cos(b)(3(sin(b))^2+(cos(b))^2)  /  −(sin(2b)) ^2 +1  </em>

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2 years ago
What is the solution set to the equation (2x−4)(4x−5)=0?
STALIN [3.7K]
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Explanation:
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