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Snowcat [4.5K]
1 year ago
7

A bug is moving back and forth on a straight path. The velocity of the bug is given by vt=t2-3t. Find the average acceleration o

f the bug on the interval 1, 4.
Mathematics
1 answer:
g100num [7]1 year ago
3 0

Answer:

2 unit/time²

Step-by-step explanation:

Given the equation:

v(t) =t^2-3t

At interval ; 1, 4

V(1) = 1^2 - 3(1)

V(1) = 1 - 3

V(1) = - 2

At t = 4

V(4) = 4^2 - 3(4)

V(4) = 16 - 12

V(4) = 4

Average acceleration : (final - Initial Velocity) / change in time

Average acceleration = (4 - (-2)) ÷ (4 - 1)

Average acceleration = (4 + 2) / 3

Average acceleration = 6 /3

Average acceleration = 2

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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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