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Umnica [9.8K]
4 years ago
10

PLEASE HELP ME...I WILL GIVE YOU BRAINLIEST. PPPPPPPLLLLLLLEEEEEAAAAAAASSSSSSEEEEEEE HHHHHHHEEEEEELLLLLPPPP MMMEE!!!!!!!!!!!!!!!

!!!!!!!!
#1. Use milligrams to prove that 1.989 . 10^27 kilograms is equal to 1.989 · 10^33 milligrams.
#2. Use kilograms to prove that 1.989 - 10^33 milligrams is equal to 1.989 . 10^27 kilograms.​
Mathematics
2 answers:
xeze [42]4 years ago
8 0

Answer:._.

Step-by-step explanation:

Ratling [72]4 years ago
7 0

Answer:

Since 1 milligram is \frac{1}{10^{6} } or 10^{-6} kilogram we can then use ratio to solve.

Step-by-step explanation:

#1)

1  :  10^{-6}

x :  1.989*10^27

Cross multiply to get:

10^{-6}x = 1.989*10^27

Divide both sides by 10^{-6} to get:

x = \frac{ 1.989*10^{27}}{10^{-6}} which is going to give you  1.989*10^33

Therefore 1.989*10^27 kilograms = 1.989*10^33 milligrams

#2) we can use the same procedure as (#1)

                 1  :  10^{-6}

1.989*10^33 : x

Cross multiply to get:

1.989*10^27 = x

Therefore 1.989*10^27 milligrams = 1.989*10^33 kilograms

Note: I'm not really sure if this is what you wanted, but hope I was able to help.

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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Part b

The 95% confident means that if we select 100 different samples and calculate the 95% confidence interval for each sample selected, then we will have approximately 95 out of 100 confidence intervals will contain the true mean for the parameter of interest.

Part c

If we have the same info but we want more confidence that implies that the confidence interval would be wider, since the margin of erroe increase with more confidence, because the critical value increase.

Part d

We need to take in count that the margin of error is given by:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}

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