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BARSIC [14]
3 years ago
5

There are 9 marbles representing 3 different colors. Write a problem where 2 marbles are selected at random without replcement a

nd the probabilty is 1/6
Mathematics
1 answer:
Nata [24]3 years ago
8 0
There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.

Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).

In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.

Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.

Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.

The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?
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Dylan invested $20,000 in an account paying an interest rate of 5.3% compounded continuously. Assuming no deposits or withdrawal
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Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Sea-hawks when they won Su
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No. These measures show a thinner team of NFL players according to the mean, variance, standard deviation, and quartiles.

Step-by-step explanation:

1) The measures of variation, namely The Range, Variance, Quartiles, Interquartiles, Sum of Squares, etc. shows us how the data are dispersed.

The Range Δ is calculated:

\Delta =305-189=116 Maximum value - Minimum value for weight

Mean:

\bar{x}=\frac{189+254+235+225+190+305+202+190+252+305}{11}\approx 231.09 \:pounds

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The Standard Deviation of the sample

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2) Since there is no preceding exercise, the comparison was made to a recent study in which a NFL player average weight is about 245 pounds (average),

Since 25% of this list are player whose weight is 192.5 lbs and 50% (2nd Quartile) =225 lbs , finally only at the 3rd Quartile we have players above the regular NFL average with 253. This, added with the other data, allow us to say that this list is not a typical of all NFL players.

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3 years ago
True or False, the following numbers are in order from greatest to least:<br> 11/10, 8/7,-2.34*
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The equation through the point (5,3) and is parallel the equation y=2/5x+7​
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Point - Slope Form: (y - 3) = 0.4(x - 5)

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Well if it's parallel, they have the same slope

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I'll do it in two forms:

Point - Slope Form: (y - 3) = 0.4(x - 5)

Slope - Intercept Form: y = 0.4x + b

3 = 2 + b

b = 1

y = 0.4x + 1

I might be wrong

3 0
3 years ago
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