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sweet-ann [11.9K]
3 years ago
14

While on vacation, Kayla goes for walks 3 times a day. Her morning walk is 1 1/2 times the distance of her afternoon walk. Her e

vening walk is 1/2 the distance of her afternoon walk. At the end of the day, she walked a total of 6 miles. Did she walk for 2 miles in the afternoon?
Mathematics
1 answer:
natta225 [31]3 years ago
6 0

Answer: 2 miles

Step-by-step explanation:

Afternoon-x; morning- 1.5x; Evening: 1/2x

Make a equation: 1.5x+x+1/2=6

2.5x+1/2x=6

5/2x+1/2x=6

6x=12

x=2

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Answer:

The explicit formula for this geometric sequence is 27.3(n-1)

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For every 5 coins carmen gets ,she gives 2 to her brother Frankie.If carmen has 9 coins,howmany coins does Frankie have?
FrozenT [24]

Answer:

Frankie have 6 coins

Step-by-step explanation:

For every 5 coins Carmen gets ,she gives 2 to her brother Frankie

Out of 5 coins, Carmen given 2 to her brother Frankie

So Carmen has 3 coins and Frankie has 2 coins

Ratio of coins , Frankie to Carmen is

2: 3

We need to find the number of coins Frankie has if Carmen has 9 coins

ratio of Frankie to Carmen is x : 9

Now make a proportion and solve for x

\frac{2}{3} =\frac{x}{9}

Cross multiply it

2*9 = 3*x

18 = 3x

Now divide both sides by 3

so x= 6

Frankie have 6 coins

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3 years ago
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

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3 years ago
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Step-by-step explanation:

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