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Usimov [2.4K]
3 years ago
8

F(x)=x^3+6x^2+x^1/2 and g(x)=x^1/2 find f(x)/g(x)

Mathematics
1 answer:
Simora [160]3 years ago
6 0
Assuming your f(x) = x^3 +6x^2 +x^(1/2) and g(x) = x^(1/2), you have
.. f(x)/g(x) = x^(5/2) +6x^(3/2) +1


____
If an exponent has any arithmetic in it, parentheses are needed around it.
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Answer:

Step-by-step explanation:

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Which of the following properties completes the proof?
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The subtraction property of inequality can be used to proof that x = 7 from  x + 8 = 15

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given that x + 8 = 15

Hence:

x + 8 - 8 = 15 - 8 (subtraction property of inequality)

x = 7

The subtraction property of inequality can be used to proof that x = 7 from  x + 8 = 15

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The length of a hypotenuse, line segment GH and triangle GJH measures 6 centimeters. A line segment JH measures 2 cm which is th
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Step-by-step explanation:

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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.  

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Answer:

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Step-by-step explanation:

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