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BabaBlast [244]
3 years ago
5

35,600 is 8.3% of what number

Mathematics
1 answer:
Over [174]3 years ago
7 0
This question can be solved by:
\frac{8.3}{100} =  \frac{35,600}{x}

Cross multiply to get: 8.3x = 3,560,000

To let the x stand along (since you are finding the x value), divide 8.3 to both sides of the equal sign.

\frac{8.3x}{8.3} =  \frac{3560000}{8.3}

Simplify to get the answer.
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Answer:

1 3/20

Step-by-step explanation:


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Andre is setting up rectangle tables for a party. He can fit 6 chairs around a single table. Andre lined up 10 tables end to end
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If there is one table (t=1) then 6 chairs (c=6) can be placed around the table, 2 along the length on each side and 1 at each end.
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Anyone good at trigonometry?<br> I need a tutor or something cause I suck
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Step-by-step explanation:

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3 years ago
g The downtime per day for a computing facility has mean 4 hours and standard deviation 0.9 hour. What assumptions must be true
Sedaia [141]

Answer:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

7 0
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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