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Naddik [55]
3 years ago
9

Need help in solving this problem. Section is under limits at infinity, horizontal asymptotes...

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0
 <span>(a) you can show the concentration by: 

amount of salt / amount of water 


amount of salt 

= 30g / L 

= [ 30 * 25g ] / [ 25L ] , 25L = 1 min 

= [ 30 * 25g ] / min 


amount of water 

= 5000 + 25t 


[ 30 * 25t ] / [ 5000 + 25t ] 

= [ 30 * 25t ] / 25*[ 200 + t ] 

= 30t / [ 200 + t ] yeah 


(b) 


lim t->∞ [ 30t / ( 200 + t ) ] 

= lim t->∞ [ ( (1 / t) * 30t ) / ( (1 / t) * ( 200 + t ) ) ] 

= lim t->∞ [ 30 / ( (200 / t) + 1 ) ] 

= 30 / ( (200 / ∞) + 1 ) 

= 30 / 1 

= 30


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
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Answer:

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Step-by-step explanation:

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Answer:

99% CI:

-4.637\leq\mu_v-\mu_o\leq 3.737

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

s_{Md}=\sqrt{\frac{2MSE}{n_h}}

The MSE can be calculated as

MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65

The weighted sample size nh can be calculated as

n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16

The standard deviation then becomes

s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

\Delta M=M_v-M_o=5.10-5.55=-0.45

Then the confidence interval can be constructed as

\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737

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2 years ago
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