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Mathematics

1 answer:

ser-zykov [4K]3 years ago

4 0

A, in one day there are 6 clients and for each day multiplied it equals C for example if you were to plug in 1 for d
6(1)=6 and so on which shows the total clients.

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Suppose that a cyclist began a 364 mi ride across a state at the western edge of the state, at the same time that a car traveli

maw [93]

The car's average rate is 43.7 mph. The bicycle's average rate is 12.3 mph.

To get this solution, let the bicycle's average rate be (x) and the car's average rate be (x + 31.4) mph.

Distance= Speed x Time
364 = (x + x + 31.4) 6.5
364 = (2x + 31.4) 6.5
364/6.5 = 2x + 31.4
56 = 2x + 31.4
56 - 31.4 = 2x
24.6 = 2x
24.6/2 = x
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x = 12.3 mph
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3 years ago

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Alenkasestr [34]

Answer:

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Oksanka [162]

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3 years ago

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3 years ago

Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings

Goshia [24]

Answer:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

6 0

3 years ago