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seropon [69]
3 years ago
6

Find the means of these numbers 3,7,2,9,4,7,3,7,5,2

Mathematics
2 answers:
Sholpan [36]3 years ago
6 0

The "mean" of a group of numbers is also called their "average".

To calculate the mean, add up all the numbers, then
divide the sum by the number of items on the list.

Step #1:  Addum up:  3+7+2+9+4+7+3+7+5+2 = 49

Step #2:  Count the number of items on the list:  I count 10 .

Step #3:  Divide the sum by the number of items on the list.

                 49 / 10  =  4.9

The mean (average) of the numbers on that list is  4.9 .

arlik [135]3 years ago
4 0
First to find the mean of the numbers we have to add the all up
3 + 7 + 2 + 9 + 4 + 7 + 3 + 7 + 5 + 2 = 49
Then we have to divide by how many numbers we have which is 10
49 divided by 10 equals 4.9
4.9 is the mean of the data set.
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Step-by-step explanation:

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3 years ago
Each month cameron budgets $1040 for fixed expenses $980 for living expenses and $120 for annual expenses his annual net income
sineoko [7]
Let us first find how much he is spending every year, to do this let's find his monthly expense and multiply it by 12.


Every month Cameron spends 1040 + 980 + 120 = 2140

To find out how much he spends yearly, multiply the monthly value by 12,
2140 x 12 = 25680


This value is more than his net income so he clearly has a surplus, but to check we can subtract 129 from every month to get:

(1040 + 980 + 120 - 129) = 2011

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5 0
3 years ago
Read 2 more answers
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Art [367]
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8 0
3 years ago
What is √49 times 7?
krek1111 [17]
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8 0
3 years ago
Read 2 more answers
Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9
nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B  

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = \left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]  = 1(1-4) - 2(2-6) + 3(4 - 3)

                    = 1(-3) - 2(-4) + 3(1)

                    = -3+8+3 = 8

also D\neq 0

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx =  \left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]   = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

x = \frac{Dx}{D}  = \frac{16}{8} = 2

similarly

Dy = \left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right] = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0

y = \frac{Dy}{D} = 0

Dz= \left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right] = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24

z = \frac{Dz}{D} = \frac{24}{8} = 3

so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

3 0
3 years ago
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