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krek1111 [17]
4 years ago
14

Identify all of the real fifth roots of 1024.

Mathematics
2 answers:
V125BC [204]4 years ago
8 0

Answer:

Real fifth root of 1024 is 4.

Step-by-step explanation:

We have to find the value of real fifth root of 1024.

For this let fifth root of 1024 = x

Therefore x^{5} = 1024 = 4^{5}

Or x^{5}=4^{5}

Therefore x = 4

X= 4 is the right answer.

muminat4 years ago
4 0

Answer:

\large\boxed{\sqrt[5]{1024}=4}

Step-by-step explanation:

\begin{array}{c|c}1024&2\\512&2\\256&2\\128&2\\64&2\\32&2\\16&2\\8&2\\4&2\\2&2\\1\end{array}\\\\1024=\underbrace{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}_{10}=2^{10}\\\\\sqrt[5]{1024}=\sqrt[5]{2^{10}}\\\\\text{Use}\ (a^n)^m=a^{nm}\\\\=\sqrt[5]{2^{2\cdot5}}=\sqrt[5]{(2^2)^5}=\sqrt[5]{4^5}\\\\\text{Use}\ \sqrt[n]{a^n}=a\\\\=\boxed4

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In a random sample of 400 items where 84 were found to be​ defective, the null hypothesis that 20​% of the items in the populati
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Answer:

z=\frac{0.21 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=0.5  

p_v =2*P(Z>0.5)=0.617  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the true proportion of defectives it's not significant different from 0.2.  

Step-by-step explanation:

1) Data given and notation

n=400 represent the random sample taken

X=84 represent the number of items defective

\hat p=\frac{84}{400}=0.21 estimated proportion of defectives

p_o=0.2 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.2 or 20%:  

Null hypothesis:p=0.2  

Alternative hypothesis:p \neq 0.2  

When we conduct a proportion test we need to use the z statisitc, 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 \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

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

z=\frac{0.21 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=0.5  

4) 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=0.1. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z>0.5)=0.617  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the true proportion of defectives it's not significant different from 0.2.  

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3 years ago
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