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nikklg [1K]
2 years ago
7

PLZ HELPPPPPPPPPPPPPPPP

Mathematics
2 answers:
Sedbober [7]2 years ago
5 0
I believe it’s 9x+1

Add all the equations/side values together
skelet666 [1.2K]2 years ago
3 0
9x+1 is the answer for this problem
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Select all the expressions with a value of less than one.
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Sjdjsksidhehshdheehrhh

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3 years ago
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Stephanie receives a salary of 650 per month plus a commission of 5.5%on the first 3000 of sales and 7% of all sales over 3000 w
Fynjy0 [20]
9290 x 7% = 650.30
650.30 + 650= 1300.30
5 0
2 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
PLEASE IM IN CLASS RIGHT NOWWWW
agasfer [191]

Answer:

102

Step-by-step explanation:

8 0
2 years ago
2 x <img src="https://tex.z-dn.net/?f=%5Csqrt%7B36%7D" id="TexFormula1" title="\sqrt{36}" alt="\sqrt{36}" align="absmiddle" clas
Andrei [34K]
The answer is 48
2x6+36
12+36=48
4 0
2 years ago
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