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kvasek [131]
2 years ago
8

How can you use bar diagrams to show a percent of change?

Mathematics
2 answers:
WINSTONCH [101]2 years ago
8 0

Answer:

Step-by-step explanation: here is a video that you can watch, that will answer your question (Solve Percent Problems Using a Tape Diagram (Bar Diagram) ) and search this

a_sh-v [17]2 years ago
7 0
Add and subtract hope this helps
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David has a garden with dimensions 11 feet long x 9 feet
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Add 3 feet to the length

Step-by-step explanation:

14 times 9 is 126

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Please Help asap alot of points
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A recipe uses 1 ¼ cups of milk to make 10 servings. If the same amount of milk is used for each serving, how many servings can b
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128 is your answer.

Step-by-step explanation:

1 gallon of milk will make 128 servings.

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When constructing parallel lines , how can you verify the lines constructed are parallel
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  compare the distance between the lines at one point to the distance between the lines at another point

Step-by-step explanation:

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3 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.

8 0
3 years ago
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