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Basile [38]
3 years ago
10

Which reasoning best supports his conclusion?

Mathematics
1 answer:
____ [38]3 years ago
5 0

Answer: You're correct! (A)

Step-by-step explanation:

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wariber [46]
1 and 2 are equations and 3 is a solution
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3 years ago
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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
Gnom [1K]

Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

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

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

5 0
3 years ago
A domain of [-4, -2, 1​
UNO [17]

what???????

???????????????????????

5 0
3 years ago
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
Could someone check this or correct me please?
Greeley [361]
I agree. !!!!!!!!!!!!!!!!!!!!!!!!
3 0
3 years ago
Read 2 more answers
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