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Juli2301 [7.4K]
3 years ago
13

What is the solution to the equation 1/2n=7

Mathematics
2 answers:
Nuetrik [128]3 years ago
5 0
1/2xn=7

Since 1/2 is being multiplied by and we have to do the inverse of multiplication which is division to get rid of n.

1/2xn/1/2=n

7/1/2=14

n=14
Nana76 [90]3 years ago
3 0
I think would equal 14, so n = 14
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The data below are the frequency of cremation burials found in 17 archaeological sites. a. Obtain the​ mean, median, and mode of
Sergeeva-Olga [200]

Answer:

(a) The mean of the data is 275.

(b) The median of the data is 85.

(c) The mode of the data is 45.

(d)  The measure of center that works best​ here will be median.

Step-by-step explanation:

We are given below the frequency of cremation burials found in 17 archaeological sites. Arranging those in <u>ascending order</u> we get;

28, 31, 32, 45, 45, 47, 59, 67, 85, 86, 143, 256, 272, 390, 424, 524, 2141.

(a) The formula for calculating mean for the above data is given by;

     Mean, \bar X  =  \frac{\sum X}{n}

                =  \frac{28+ 31+ 32+ 45+ 45+ 47+ 59+ 67+ 85+ 86+ 143+ 256+ 272+ 390+ 424+ 524+ 2141}{17}

                =  \frac{4675}{17}  = 275

So, the mean of the data is 275.

(b) Since, the number of observations (n) in our data is odd (i.e. n = 17) , so the formula for calculating median is given by;

                    Median  =  (\frac{n+1}{2} )^{th} \text{ obs.}

                                   =  (\frac{17+1}{2} )^{th} \text{ obs.}

                                   =  (\frac{18}{2} )^{th} \text{ obs.}

                                   =  9^{th} \text{ obs.} = 85

So, the median of the data is 85.

(c) <u>Mode</u> is that value in our data which appears maximum number of times in our data.

So, after observing our data we can see that only number 45 is appearing maximum number of times (2 times) and all other numbers are appearing  only once.

So, the mode of the data is 45.

(d) The measure of center that works best​ here will be median because there are outliers in our data (means extreme values) and mean gets affected by the outliers.

So, the best measure would be median as it represents the middle most value of our data.

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