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3 years ago

Two numbers have a sum of 117 and a difference of 43. Find the numbers.

Mathematics

2 answers:

blondinia [14]3 years ago

6 0

Answer:

this kid grayson is deleting my answer and i dont know how when they arent even his questions and he is being a little brat

Step-by-step explanation:

so watch out for him

Elza [17]3 years ago

4 0

The answer is B. Please mark brainliest! Have a great day!

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Step-by-step explanation:

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Alvin is 11 years younger than Elga. The sum of their ages is 43. What is elgas age

Neporo4naja [7]

Answer:

Step-by-step explanation:

Lets assume x is Elgas' age

And y is Alvins age

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x=y+11

and the sum of their age is 43

x+y=43

solve it and you get 16

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3 years ago

Graph the solution of y-2>1 on a number line

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3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

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

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago