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kumpel [21]
3 years ago
15

Mitchell has a small bag with 12 M&Ms in it. 5 of the M&Ms are yellow. 2 are green. 1 is orange. 4 is brown. what fricti

on of the M&M are brown?​ get BRAINLIEST
Mathematics
1 answer:
Sophie [7]3 years ago
6 0

Answer:

1/3

Step-by-step explanation:

There are 4 brown M&Ms and 12 M&Ms in total.

Since both 4 and 12 are divisible by 4,

4/12 = 1/3

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Here is the expansion of <span>(x + y)n </span><span>for </span><span>n = 0, 1,…, 5</span><span> :</span>

<span>(x + y)0 = 1</span> 
<span>(x + y)1 = x + y </span>
<span>(x + y)2 = x 2 +2xy + y 2 </span>
<span>(x + y)3 = x 3 +3x 2 y + 3xy 2 + y 3 </span>
<span>(x + y)4 = x 4 +4x 3 y + 6x 2 y 2 +4xy 3 + y 4 </span>
<span>(x + y)5 = x 5 +5x 4 y + 10x 3 y 2 +10x 2 y 3 +5xy 4 + y 5 
</span>
hopefuly this helps you understand


7 0
3 years ago
Select each situation that models a constant rate of change
nekit [7.7K]
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5 0
3 years ago
How do you graph for the function Y= -x+1
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5 0
3 years ago
Read 2 more answers
How much can Sophia spend on party favors
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8 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
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