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Snezhnost [94]
3 years ago
8

Which compares the modes of the data? The mode for the students is 2 and the mode for the teachers is 4. The mode for the studen

ts is 0 and the mode for the teachers is 1. The mode for the students is 4 and the mode for the teachers is 8. The mode for the students is 2 and the mode for the teachers is 1.
Mathematics
2 answers:
Goshia [24]3 years ago
6 0

Answer:

the right answer is b.The mode for the students is 0 and the mode for the teachers is 1.

Step-by-step explanation:

hope it works

motikmotik3 years ago
5 0

Answer:

B

Step-by-step explanation:

DID THE WORK

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12 31/100 is the mixed fraction because in a decimal all the numbers to the left are whole numbers and all the numbers to the right are numbers that are part of a whole. Just like in a mixed fraction there is a whole number and a fraction in a mixed number. Since 12 is in the left it is the whole number and since 31 is to the right and in the place value 31 is ending in the hundredths place so that's why the fraction is, 31/100. That's how you came to know that 12 31/100 is the fraction for 12.31.
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There are 6 cups of sugar in 5 batches of cookies. how many cups are in 1 batch
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Step-by-step explanation:

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Simplify the following radical expressions. 100 points<br> √8/√6<br> √2/10<br> √532<br><br> √3k√8
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Step-by-step explanation:

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3 years ago
The shoes were $175. They are now 10% off. How much are they now?
kondor19780726 [428]

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Hope this helps!

3 0
3 years ago
Suppose that you are in charge of evaluating teacher performance at a large elementary school. One tool you have for this evalua
Strike441 [17]

Answer:

a) Standard error = 2

b) Range = (76.08, 83.92)

c) P=0.69

d) Smaller

e) Greater

Step-by-step explanation:

a) When we have a sample taken out of the population, the standard error of the mean is calculated as:

\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5

Then, the probability of getting a average score of 79 or higher is:

P(X>79)=P(z>-0.5)=0.69146

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

d) If the sample is smaller, the standard error is bigger (as the square root of the sample size is in the denominator), so the spread of the probability distribution is more. This results then in a smaller probability for any range.

e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

6 0
3 years ago
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