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Elenna [48]
3 years ago
13

Evaluate 32 + (5 − 2) ⋅ 4 − 6 over 3

Mathematics
2 answers:
Ann [662]3 years ago
6 0

Answer:

12.66666 infinity or 38/3

Step-by-step explanation:

Parentheses first so 5-2 is 3... then multiply so then 12... 12+32-6 is 38. 38 over 3 is 12.6666 infinity or 38/3, cannot be simplified...

Masteriza [31]3 years ago
3 0
The answer is 38/3 or a different form of the Answer could be 12 2/3
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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

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The mean calculated for this case is \bar X=0.475

The sample deviation calculated s=0.183

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Null hypothesis:\mu = 0.5  

Alternative hypothesis:\mu \neq 0.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=8-1=7  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

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