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svet-max [94.6K]
2 years ago
12

A jar contains three marbles; 1 yellow (y), 1 orange (o), and 1 red (r). Which list shows all the possible outcomes for choosing

two marbles at random from the jar if the first marble is replaced after it is drawn?
How many of the outcomes contain at least one yellow marble?
Mathematics
2 answers:
Zarrin [17]2 years ago
4 1

Answer:

see below

Step-by-step explanation:

Since we replace the marble , we can draw the same color again

yy    oo     ro

yo    oy     ry

yr     or      rr

There are 9 possible outcomes for drawing 2 marbles

5 of them have at least one yellow marble

Guest
1 year ago
kinda thanks cause you only helped me with part of this
VLD [36.1K]2 years ago
4 1

Answer:

there are 9 possible outcomes

Step-by-step explanation:

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Answer:

0. 4329

Step by step explanation

Since both groups will have the same number of men, we would guarantee each group consists of three women and three men.

So theprobability is (6, 3)·(6, 3)/(12, 6)=100/231.

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Assume that the number of apples is x and the number of oranges is y.

For the first given, we know that each apple costs $0.24 and each orange costs $0.8, therefore:
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we also know that the total amount spent is $12, therefore the first equation is as follows:
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x + y = 20

You can easily graph these two functions and find a possible combination from the graph (the correct combination would be the intersection between the two lines).


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An artist wants to make alabaster pyramids using a block of alabaster with a volume of 576 cubic inches. She plans to make each
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Hank bought 2.4 pounds of apples . Each pound cost $1.95. How much did hank spend ont the apples ?
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The first task of the Segment Two Honors Project is to select the Power Pill study participants’ groups and research board offic
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Step-by-step explanation:

Part 1:

The number of ways in which we can organized n elements into k groups with size n1, n2,...nk is calculate as:

\frac{ n!}{ n1!*n2!*...*nk! }

So, in this case we can form 4 subgroups with 10 participants each one, replacing the values of:

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We get:

\frac{ 40!}{10!*10!*10!*10!} = 4.7*10^{21}

Part 2:

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