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3 years ago

Paul has 30$ to spend at the fair if the admission to the fair is 5$ and the rides cost 3.50 each how many rides can paul go on

2 answers:

const2013 [10]3 years ago

5 0

by the time Paul got into the fair he would have already paid admission with his 30 bucks meaning that he would have 25 dollars left for rides and 25 divided by 3.50 is 7.blah blah blah the rest dosent matter so Paul could go on a maximum of 7 rides

Margaret [11]3 years ago

4 0

Answer:

Step-by-step explanation:

Start with $30.

Subtract $5 for the admission: $30 - $5 = $25.

Now he has $25 to spend on rides. Each ride costs $3.5, so we divide $25 by $3.50.

$25/$3.50 = 7.14

Since the number of rides must be a whole number, he can go on 7 rides.

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Thirty percent of all customers who enter a department store will make a purchase. Suppose that 6 customers enter the store and

torisob [31]

Answer:

Step-by-step explanation:

Let's use binomial distribution, because we are interested in find the total number of successes in a sequence of n independent experiments.

In probability, a binomial distribution is used in a Bernoulli process. That is, a sequence of experiments of n independent trials, each asking a dichotomous question, that means it only has two answers. One answer is a success (probability: p) and the other is a failure (probability: 1-p). In this case, every customer who enters in the store is the experiment. If they make a purchase is a success event, taking into account that every purchase decision has to be independent.

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

x= total number of success

n= total number of experiments

p=probability of success on an indivual trial

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

a) x= 5 ------> customers that make a purchase

n= 6 ------> total customers

p=0.3

$p(X=5)= C(6,5)(0.3)^{5} (1-0.3)^{6-5} \\p(X=5)=0.102$

b) At least 3 customers make a purchase. x ≥ 3

$P(X\geq 3)=1-P (X=0)-P(X=1)-P(X=2)\\P(X\geq 3)=1-(C(6,0)(0.3^{0})(0.7)^{6}) - (C(6,1)(0.3^{1})(0.7)^{5}) - (C(6,2)(0.3^{2})(0.7)^{4})\\ P(X\geq 3) = 1-0.11765-0.30253-0.32414\\P(X\geq3)= 0.2557$

c) At most 2 customers make a purchase. x ≤ 2

$P(X\leq2) = P(X=0) + P(X=1) + P(X=2)\\P(X\leq2) = (C(6,0)(0.3^{0})(0.7)^{6}) + (C(6,1)(0.3^{1})(0.7)^{5}) + (C(6,2)(0.3^{2})(0.7)^{4})\\P(X\leq2) = 0.11765-0.30253-0.32414\\P(X\leq2)=0.7443$

d)At least 1 customer makes a purchase. x ≥ 1

$P(X\geq 1)=1-P(X=0)\\P(X\geq 1)= 1-C(0,6)(0.3)^{0}(0.7)^{6} \\P(X\geq 1)= 1-0.11765 = 0.8824$

e)They must be kind to customers, always willing to help and give information about products clearly and honestly.

Have enough inventory and offer promotions for the purchase of more than one product

Have an agile payment system that avoids rows and delays in purchases.

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VLD [36.1K]

Answer:

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Step-by-step explanation:

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2 years ago