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Talja [164]
3 years ago
9

Need help ...............................

Mathematics
1 answer:
vekshin13 years ago
7 0
The answer is region 1.
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....sigh....<br> Shout out to Taater for being helpful in times of need :3
Ne4ueva [31]

Answer:

Taater is a good friend of mine.

Step-by-step explanation:

The answer would be the third option, hope I'm just as helpful in times of need if its possible to subtract those then it's 100% the first option. Sorry don't exactly know this, so I'm at least trying to help.

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3 years ago
Challenge A movie theater sends out a coupon for 30% off the price of a ticket. Write an equation
BartSMP [9]

Answer:

The equation is y= 0,65 x

Step-by-step explanation:

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2 years ago
Do anyone know how to answer this?
ANTONII [103]

Answer:

10

Step-by-step explanation:

8 0
3 years ago
3. approximate value of log 3 base 7​
xxTIMURxx [149]

Answer:

0.56457

Step-by-step explanation:

log 3  with base 7  =    (Log 3)/ (Log 7) = 0.47712125 / 0.845098040014

(Log 3) / (Log 7) = 0.56457

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3 years ago
If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distributio
goldenfox [79]

Answer:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And we want a normal distribution for the sample mean

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

X_i \sim N(\mu , \sigma), i=1,2,...,n

Because if we find the parameters we got:

E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

And the deviation would be:

Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

And we satisfy the condition:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

3 0
2 years ago
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