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vodka [1.7K]
3 years ago
9

A club has 25 members. How many ways are there to choose four members of the club to serve on the executive committee? How many

ways are there to choose a president, vice-president, secretary, and treasurer of the club
Mathematics
1 answer:
Contact [7]3 years ago
4 0

Answer:

303600 ways

Step-by-step explanation:

No. of members = 25

We are supposed to find the no. of ways to choose a president, vice-president, secretary, and treasurer of the club

No. of posts = 4

The order matters here since each member will get different posts .

So, we will use permutation over here

Formula: ^nP_r=\frac{n!}{(n-r)!}

n = 25

r = 4

^{25}P_4=\frac{25!}{(25-4)!}

^{25}P_4=303600

Hence there are 303600 ways to choose a president, vice-president, secretary, and treasurer of the club

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the sum of two numbers is 100. the sum of 8 and the larger number is equal to 5 times the smaller number. What is the smaller nu
crimeas [40]

Answer:

x=82

y=18

Step-by-step explanation:

x+y=100

8+x=5y

5 0
3 years ago
We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
777dan777 [17]

Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

8 0
3 years ago
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Novosadov [1.4K]

Answer:

30

Step-by-step explanation:

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Divide

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Answer:

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8 0
3 years ago
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