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skad [1K]
3 years ago
6

I only need help on the one that are mark blue

Mathematics
1 answer:
Oxana [17]3 years ago
4 0
1. (54/9)6 equation
     6 x 6
       36 answer
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What is 2 divided by 1386
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Lol its <span>0.00144300144</span>
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3 years ago
In which quadrant is point C located
kozerog [31]

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Quadrant 4

Step-by-step explanation:

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The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w
dimulka [17.4K]

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

8 0
3 years ago
This is pretty easy :0
maksim [4K]

Answer:

1, 1.5, -1 1/5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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