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avanturin [10]
3 years ago
7

A decimal with a repeating digit of 0

Mathematics
1 answer:
natima [27]3 years ago
6 0

Answer:

Necessarily it could be anything

Step-by-step explanation:

If you want a repeating decimal digit of 0 it's going to be all decimals just to a different place like the tenths place. For example 0.5 is also 0.500000000000... That is why pretty much every decimal has a repeating digit of 0.

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Step-by-step explanation:

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Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give t
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Answer:

P(\bar X >80)=P(Z>2.143)=1-P(z

Step-by-step explanation:

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".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

X \sim N(\mu=74, \sigma=14)

The sampling distribution for the sample mean is given by:

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

The deduction is explained below we have this:

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

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

Since the variance for each individual observation is Var(x_i)=\sigma^2 then:

Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}

And then for this special case:

\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)

We are interested on this probability:

P(\bar X >80)

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

Applying this we have the following result:

P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)

And using the normal standard distribution, Excel or a calculator we find this:

P(Z>2.143)=1-P(z

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Step-by-step explanation: since i helped can i have brainlist please :D

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