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NARA [144]
3 years ago
9

What is 5/6 of 15 in simplest form.

Mathematics
1 answer:
aksik [14]3 years ago
6 0

12  1/2 because you do 5/6 x 15/1 which 6/3 equals 2 and 15/3 equals 5 so you do 5/2 x 5/1

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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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3 years ago
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A tortoise and a hare are competing in a race around a 1600-meter track. The arrogant hare decides to let the tortoise have a 78
Sindrei [870]
I am presuming that the question is who won or by how much did the winner win. 

distance equals rate times time, or d = r*t
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The hare finished in 160 seconds

For the tortoise d = 1600 - 780 = 820 (due to the head start) and r = 5.3
820= 5.3 t
Divide both sides by 5.3
t = 154.72 (rounded to the hundredths place)

The tortoise won by 5. 28 seconds (160-154.72)
 
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Step-by-step explanation:c 15

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