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Delvig [45]
3 years ago
6

The U.S. Mint began issuing state quarters in 1999. There was one state quarter for each of the 50 states. You are collecting th

e state quarters and want to design a rectangular display with the same number of quarters in each row. How many ways can you arrange your display?
Mathematics
2 answers:
7nadin3 [17]3 years ago
6 0

Answer:

50

Step-by-step explanation:

If we have 50 states, and that is a even number. that means we can make rectangular shapes with nothing leftover. They can even get to being one line thick. ................... So it can be aranged in 50 ways.

ddd [48]3 years ago
3 0

Answer:

probably 100-200

Step-by-step explanation;

that is a hard question to be honest

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A salesman has a base salary $2600 per month and makes 3 commission on each unit sold. If his total annual compensation is $48,0
nevsk [136]

Answer:

Each unit cost $800

Step-by-step explanation:

Annual compensation = $48,000

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4 years ago
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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.

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luda_lava [24]
The answer is x= -5 ...explained... -2x + 5 -35x= 190...-37x +5 = 190 ... now subtract 190-5 = 185... divide 185 by - 37 and get -5
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