$1799.91 with 6% tax added

2 answers:

6 0

To find tax : $1,799.91 x 1.06 = $1907.9046. Because it's money, it should be rounded to $1907.90.

Of course, another way to do it is by multiplying $1,799.91 by 0.06, and adding the product that you get to 1,799.91.

Either way, you will still get $1907.9046

Harrizon [31]3 years ago

6 0

1799.91 with 6%

First lets make 6% into a decimal

6% divided by 100 = 0.06
6% = 0.06

Now lets multiply

1799.91 x 0.06 = 107.9946

Now lets add

107.9946 + 1799.91 = 1,907.9046

The final answer is 1,907.9046

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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 X = number of orange milk chocolate M&M’s.

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

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

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

The random variable X follows a Binomial distribution with parameter n = 55 and p = 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:

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