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wariber [46]
1 year ago
15

You get a $60,000 loan for a new dump truck. With the truck, you can generate $500 more business each week at 80% margin. Your i

nterest rate on the loan is 8.6%. Does the additional margin generated cover the Interest payments?​
a) Yes
b) No
Mathematics
1 answer:
Levart [38]1 year ago
4 0

Answer: Yes

Step-by-step explanation: got it wrong and told me the correct answer

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According to Eurostat, 55.2% of households in Italy have one child, the highest percentage in the Euro zone. A random sample of
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0.04

Step-by-step explanation:

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SE= \sqrt{\frac{p *(1-p)}{n} } \\

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A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr
Gre4nikov [31]

Answer:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:P(-1.5 \leq Z \leq 1.5) = P(zStep-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(50,12)  

Where \mu=50 and \sigma=12

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

We can find the probability required like this:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:

P(-1.5 \leq Z \leq 1.5) = P(z

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