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Ksenya-84 [330]
3 years ago
7

What is the difference b/w "Binomial Series" and "Binomial Theorem"? Also give formula for each.

Mathematics
1 answer:
horrorfan [7]3 years ago
8 0

Step-by-step explanation:

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The binomial theorem states that for some a,b∈R and some k ∈Z+ ,

(a+b)k=∑n=0k(kn)ak−nbn.

The binomial series allows us to use the binomial theorem for instances when k is not a positive integer. The binomial series applies to a given function f(x)=(1+x)k for any k∈R with the condition that |x|<1 . It is stated as follows:

(1+x)k=∑n=0∞(kn)xn .

Note that the binomial theorem produces a finite sum and the binomial series produces an infinite sum.

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Michael is buying a pair of jeans that regularly cost $40. They are on sale for 20% off. If the tax rate is 8%, what is the sale
Anna35 [415]

Answer      

Find out the  what is the sale price of the jeans including tax .

To prove

Let us assume that the sale price of the jeans including tax be x .

As given

Michael is buying a pair of jeans that regularly cost $40.

They are on sale for 20% off.

If the tax rate is 8% .

20% is written in the decimal form

= \frac{20}{100}

= 0.20

8% is written in the decimal form

= \frac{8}{100}

= 0.08

Than the equation becomes

x = 40 - 40 × 0.20 + 40 × 0.08

x = 40 - 8 + 3.2

x = 43.2 - 8

x = $35.2

Therefore the sale price of the jeans including tax be $ 35.2 .



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3 years ago
(-4+5)/(-1)+3-21/(-7)13[-11(-2)-19
sukhopar [10]

Answer:

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Step-by-step explanation:

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nika2105 [10]

Answer:

9. \sqrt{61} \sqrt{71} 8 10  

10. 6 \sqrt{45} \sqrt{63} 9

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio
yaroslaw [1]

Answer:

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.   

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion  is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

6 0
3 years ago
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