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erik [133]
3 years ago
15

Need help really bad, brainliest to who answers,

Mathematics
1 answer:
Paraphin [41]3 years ago
7 0

Answer:

I know (A, (E, and (F

Note:

There might be more.

<em>-kiniwih426</em>

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The answer is 0.40 or 0.4
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1- (-1,4)

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Given h(x) = 5x + 2, solve for a when h(x) = -8.
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The initial mass of a certain species of fish is 6 million tons. The mass of​ fish, if left​ alone, would increase at a rate pro
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Step-by-step explanation:

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3 years ago
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
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