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3 years ago

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Jeanette purchased a concert ticket on the website. The original price of the ticket was $75. She used a coupon code to receive

a 20% discount. The website apply to 10% service fee to the discounted price. Jeanettes ticket was less than the original price by what percent?

2 answers:

Marizza181 [45]3 years ago

8 0

Answer: Jeanettes ticket was less than the original price by 28%

Step-by-step explanation:

The original price of the ticket was $75. She used a coupon code to receive a 20% discount. This means that the amount of discount that she got is

20/100 × 75 = 15

The discounted price would be

75 - 15 = $60

The website apply to 10% service fee to the discounted price. This means that the amount of service fee charged by the website is

10/100 × 60 = $6

The final cost of the ticket is

60 - 6 = $54

The difference between the final price and the original price is

75 - 54 = 21

The percentage difference in the prices would be

21/75 × 100 = 28%

xxMikexx [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

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a) 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) 80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

c) 4.074

d) Variance 3.24, standard deviation 1.8

Step-by-step explanation:

For each internet browser user, there are only two possible outcomes. Either they use chrome, or they do not. They are chosen at random, which means that the probability of an user using chrome is independent from other users. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

20.37% share of the browser market

This means that $p = 0.2037$

Group of 20 Internet browser users

This means that $n = 20$

(a) Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is $P(X = 8)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

(b) Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

Either less than 3 users use Chrome, or at least 3 do. The sum of the probabilities of these events is decimal 1. So

$P(X < 3) + P(X \geq 3) = 1$

So

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.195$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.195 = 0.805$

80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

(c) For the sample of 20 Internet browser users, compute the expected number of Chrome users.

$E(X) = np = 20*0.2037 = 4.074$

(d) For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

$V(X) = np(1-p)$

$V(X) = 20*0.2037*0.7963 = 3.24$

The standard deviation is the square root of the variance. SO

$\sqrt{V(X)} = \sqrt{3.24} = 1.8$

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3 years ago