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algol13
3 years ago
11

Jill and Mary are selling cheesecakes for a school fundraiser. Customers can buy pecan cheesecakes and apple cheesecakes. Jill s

old 6 pecan cheesecakes and 14 apple cheesecakes for a total of $382. Mary sold 3 pecan cheesecakes and 3 apple cheesecakes for a total of $111. Find the cost each of one pecan cheesecake and one apple cheesecake.
Mathematics
2 answers:
xxMikexx [17]3 years ago
8 0

Answer:

Mary sold her cheesecakes for $18.50

Jill sold her cheesecakes for $19.10

Step-by-step explanation:

bulgar [2K]3 years ago
4 0
Jill sold her cheesecakes for $19.10 
Mary sold her cheesecakes for $18.50
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1. (10 marks) A study of the home purchasing habits of Canadians in show homes found the following interesting facts. The probab
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Answer:

1 a. P(X=1)=0.3809\\\\b. P(X\geq 1)=0.5721\\c. \mu_x=0.81, \ \ \ \sigma_x=0.7371\\d. Increase \ n

2.a. 0.2438\\b. 0.6227\\c. \mu_8=17.6, \ \sigma_8=4.1952

Step-by-step explanation:

The purchase of a house has binomial distribution with p=9% and n=9

a. To find the probability of one house being sold:

p(x)={n\choose x}p^x(1-p)^{n-x}

The value of x is 1:

p(x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X=x)={9\choose 1}0.09(1-0.09)^8\\\\=0.3809

Hence, the probability  that one house is sold that day is 0.3809

b.The probability of at least one house is the same as the 1 minus the probability that no house was sold:

p(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 1)=1-P(X=0)\\\\=1-{9\choose 0}0.09^0(0.91)^9\\\\=1-0.4279\\\\=0.5721

Hence,the probability that at least one house is sold that day is 0.5721

c. The mean of a binomial distribution is the product of the number of events times the probability of success=np

From a above, n=9 and p=0.09:

\mu_x=np, \ n=9 , \ p=0.09\\\\=9\times 0.09\\\\=0.81

#The standard deviation is calculated as np(1-p):

\sigma_x=np(1-p),\  n=9, p=0.09\\\\=9\times 0.09\times 0.91\\\\=0.7371

Hence, the distribution has a mean of 0.81 and a standard deviation of 0.7371

d. Given that the probability of success is low, the overall successful events can only be increased by increasing the sample size, say n should be increased to 1000.

2. a.Given a Poisson distribution with mean=2.2

-The probability in a Poisson distribution is expressed as:

P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}

The probability of one person showing up is calculated as:

P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}\\\\P(1,2.2)=\frac{e^{-2.2}2.2^1}{1!}\\\\=0.2438

Hence , the probability that exactly one person shows up during a 1-hour period is 0.2438

b. the probability that at most 2 people will show up during a 1-hour period is calculated as:

P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}\\\\P(x\leq2 ;\mu)=P(x=0;\mu)+P(x=1;\mu)+P(x=2 ;\mu)\\\\=\frac{e^{-2.2}2.2^0}{0!}+\frac{e^{-2.2}2.2^1}{1!}+\frac{e^{-2.2}2.2^2}{2!}\\\\=0.1108+0.2438+0.2681\\\\=0.6227

Hence,the probability that at most 2 people will show up during a 1-hour period is 0.6227

c. The mean and variance of a Poisson distribution is equal:

E(X)=\mu=Var(X)=\sigma^2, \mu=2.2\\\\8E(X)=8(2.2)=\mu_8\\\\\mu_8=17.6\\\\V(X)=\sigma^2=17.6\\\\\sigma=4.1952

Hence the mean after 8 hrs is 17.6 and the standard deviation will be 4.1952

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