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notka56 [123]
3 years ago
11

A cruise company would like to estimate the average beer consumption to plan its beer inventory levels on future​ seven-day crui

ses.​ (The ship certainly​ doesn't want to run out of beer in the middle of the​ ocean!) The average beer consumption over 1717 randomly selected​ seven-day cruises was 82 comma 35782,357 bottles with a sample standard deviation of 4 comma 4924,492 bottles. Complete parts a and b below. a. Construct a 95​% confidence interval to estimate the average beer consumption per cruise. The 95​% confidence interval to estimate the average beer consumption per cruise is from a lower limit of (blank) bottles to an upper limit of (blank) bottles.
​(Round to the nearest whole​ numbers.)

b. What assumptions need to be made about this​population?

A.

The only assumption needed is that the population size is larger than 30.

B.

The only assumption needed is that the population follows the​Student's t-distribution.

C.

The only assumption needed is that the population distribution is skewed to one side.

D.

The only assumption needed is that the population follows the normal probability distribution.
Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0
I think the correct one is C
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3 years ago
3 turtles walking.
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Divide 250 by 9.4 to get 26.6

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3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
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What is the conjugate of -3+2i
WARRIOR [948]
\text{conjugate:\ \boxed{ -3-2i}}
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