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NeTakaya
4 years ago
6

Anyone know this ? Please help :(

Mathematics
1 answer:
Akimi4 [234]4 years ago
6 0

Answer:

D

Step-by-step explanation:

Just plug in and check.  Plug in the first digit for x and the second digit for f(x)

First, try A. Does 5^0= 5?  no so A is wrong

Now try B.  Does 5^5=1?  no so B is also wrong

Next try C.  Does 5^0=0? no so C is wrong.

Finally, try D.  Does 5^1=5?  Yes so D is correct.

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The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where paramet
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Answer:

a) t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744    

The degrees of freedom are given by:

df=n-1=45-1=44  

The critical value for this case is t_{\alpha}=1.68 since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) p_v =P(t_{(44)}>2.744)=0.0044  

Step-by-step explanation:

Information given

\bar X=18.4 represent the sample mean

s=2.2 represent the sample standard deviation

n=45 sample size  

\mu_o =17.5 represent the value to verify

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:  

Null hypothesis:\mu \leq 17.5  

Alternative hypothesis:\mu > 17.5  

The statistic is given by:

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

And replacing we got:

t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744    

The degrees of freedom are given by:

df=n-1=45-1=44  

The critical value for this case is t_{\alpha}=1.68 since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

Part b

The p value would be given by:

p_v =P(t_{(44)}>2.744)=0.0044  

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