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Bess [88]
3 years ago
14

What is 312.15 in expanded form?

Mathematics
1 answer:
Andrei [34K]3 years ago
3 0
300+10+2 +.10=.5 hope this helps you
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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s
Oksi-84 [34.3K]

Answer:

a) \mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

b) From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

c) P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

P(Z\geq2.070)=1-P(Z

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

\mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

Part b

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

Part c

For this case we want this probability:

P(\bar X \geq 285)

And we can use the z score defined as:

z=\frac{\bar x -\mu}{\sigma_{\bar x}}

And using this we got:

P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

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

P(Z\geq2.070)=1-P(Z

8 0
3 years ago
Composite function is the result of using one function as the input for another function.
ipn [44]

Answer:

True

Step-by-step explanation:

That is the definition of composite function

3 0
3 years ago
When Maria finished medical school, she had a student loan. She repaid $27,000 of the loan within two years of graduating. That
Alchen [17]

Answer

$54,000

Equation

x=27,000

2x=54,000

3 0
4 years ago
Georgia’s scores in six subjects are 89, 75, 92, 68, 78, 75, and 80. If she finds a single value that summarizes all the values
posledela
It would be mean. The mean is the average that is you are looking for, since she wants a summerization of all the values in the data set. To find the mean (average) add all the numbers and then divide by the amount of numbers.
5 0
3 years ago
The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of
SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 20.8, \sigma = 4.5

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 20.8}{4.5}

Z = 0.04

Z = 0.04 has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

Z = \frac{X - \mu}{\sigma}

Z = \frac{23 - 20.8}{4.5}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 18

Z = \frac{X - \mu}{\sigma}

Z = \frac{18 - 20.8}{4.5}

Z = -0.62

Z = -0.62 has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0
3 years ago
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