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sveticcg [70]
3 years ago
11

Please help me c:

Mathematics
1 answer:
netineya [11]3 years ago
8 0

Oscar played games vs number of points he scored is, C) positive, linear association.

Step-by-step explanation:

  • no association is when points Oscar graph will remain between 8to10.
  • number of games he scored his points remain the same which is mean.
  • non linear is only when there is no straight line passing.
  • Linear is either exponential or polynomial.
  • Positive as the game increase he scoring abilities increases.
  • Negative as the game increases his scoring decreases.
  • Negative x axis will have more number of points.
  • Negative y axis will high to low of the graph.
  • Linear lines are best way to predict a data doesn't work will all data.

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

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My best bet would be d
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Inessa05 [86]
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Answer:

Step-by-step eWrite 2.9 as  

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