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34kurt
3 years ago
5

4⋅10 ​6 ​​ 4, dot, 10, start superscript, 6, end superscript is how many times as large as 1\cdot10^41⋅10 ​4 ​​ 1, dot, 10, star

t superscript, 4, end superscript?
Mathematics
1 answer:
lesya [120]3 years ago
6 0

The given problem is very confusing since it was copy pasted directly from the source so the equations look scrambled and plus it was one words. After my own translation, I believe the given numbers are:

4 ⋅ 10^6

and

1 ⋅ 10^4

The symbol ⋅ means that the two numbers are multiplied while the symbol ^ means an exponent of. Now we are asked to find how much the 1st number is larger than the 2nd number. To solve this, we simply have to divide the bigger number by the smaller number. Since 4 ⋅ 10^6 has bigger exponent than 1 ⋅ 10^4 then it is the bigger number.

Ratio = 4 ⋅ 10^6 / 1 ⋅ 10^4

Ratio = 4 ⋅ 10^2 = 400

Therefore 4 ⋅ 10^6 is 400 times bigger than 1 ⋅ 10^4.

 

Answer:

<span>400 times</span>

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The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.
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Answer:

Critical values:t_{\alpha/2}=-2.201 t_{1-\alpha/2}=2.201

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) 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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The data is:

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In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is \bar X= 4.059

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n-1=12-1=11

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for t_{\alpha/2}=-2.201

"=T.INV(1-0.025,11)" for t_{1-\alpha/2}=2.201

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval :  "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

So the 95% confidence interval would be given by (3.646;4.472)

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