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lilavasa [31]
2 years ago
10

Let's consider a jar consists of 20 marbles. 12 are red and 8 are green. John picked 2 marbles at random. What is the probabilit

y that John picked both the marbles red assuming that the first one is not replaced?
Mathematics
1 answer:
adelina 88 [10]2 years ago
4 0

Answer:   \bold{\dfrac{33}{95}}

<u>Step-by-step explanation:</u>

First pick    and      Second pick      =  Probability

    \dfrac{12\ red}{20\ total}\quad \times    \dfrac{11\ red\ remaining}{19\ total\ remaining}\\  =\dfrac{132}{380}\\  

    Simplify:

 \dfrac{132}{380}\div \dfrac{4}{4}=\large\boxed{\dfrac{33}{95}}\\

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Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

  • Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

  • Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

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Next, multiply that by the sample size of 732.

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  • For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

  • P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

  • For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

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2 years ago
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See the picture to better understand the problem

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step 3. 196x^2 - 121y^2 = (14x + 11y)(14x - 11y) works!

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