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Amanda [17]
3 years ago
6

Which is more, 6 kilograms or 5,999 grams?

Mathematics
2 answers:
Andrei [34K]3 years ago
5 0
6 kg because it is 1 gram more. there are 1000 grams per kilogram, so 5,999 grams is the same as 5.999 kg. 6 kg is larger.
vivado [14]3 years ago
5 0
6 kilograms is more than 5,999 grams.
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What is the answer for 3912.99/15.9
san4es73 [151]
The question asks the answer that comes out from the division of 3912.99 by 15.9. Both the numbers are in decimal form, so the division will be easy.
3912.99/15.9
= (391299 * 10)/(159 * 100)
= (391299)/(1590)
= 246.10
So the answer that comes out from the division of the two numbers given in the question is 246.10. I hope i have described the full process and it has helped you to understand the way these kind of problems need to be tackled.
8 0
3 years ago
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Arc is 41 and angle is 40. I have no other given information but I have to find x
tiny-mole [99]

First of all I want to point out you drew the diagram a little wrong. The Arc is 41 doesn't mean its 41 degrees it means it has length 41 so remove the degrees symbol.

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3 years ago
What is 43.586 rounded to the nearest tenth
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8 0
4 years ago
Craig has of collection of 220 marbles. He lets his friend have 14 of his collections and sells 15 of his collection to the nove
Illusion [34]
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3 0
4 years ago
A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned th
AveGali [126]

Answer:

\hat \mu = \bar X = 25000

Step-by-step explanation:

Data given

\bar X = 25000 represent the sample mean obtained

s= 2500 represent the sample deviation obtained

\mu =? represent the true mean (variable of interest)

n=400 represent the sample size selected

Solution to the problem

The best estimator for the true mean \mu is given by the sample mean \bar X, and the reason is: If we assume that in tha random sample X_1,X_2,....,X_{400} each observation follows a normal distribution like this:

X_i \sim N(\mu, \sigma)

Then when we find the expected value for the sample mean we have this:

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

If we use the properties of the expected value we have:

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

And since the expected value of each observation is the same we have:

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

So then we have that the best estimator for the population mean is given by:

\hat \mu = \bar X = 25000

7 0
3 years ago
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