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dexar [7]
3 years ago
8

Avg these 190258.50 152698.00 122753.00 220523.00 231951.00

Mathematics
1 answer:
Andre45 [30]3 years ago
6 0
To get an average, you must do the following steps.

step 1:add all the numbers together.
190258.50+152698.00+122753.00+220523.00+231951.00= 918183.5Step 2: divide the ANSWER you get by the amount of numbers there are:
(in this case, it will be 5, because you added 5 numbers together)
918183.5/5


your answer is: 183636.7
that is your average

hope this helps :D
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A cell phone company charges a monthly fee of $18 plus five cents for each call. A
aleksandr82 [10.1K]

Answer:

650 calls

Step-by-step explanation:

so since you have 18$ per month plus 5 cents per call you would do

18+0.5n(n represent the number of calls)= the total fee of $50.50 cents.

thus,now you need to figure out how much the phone calls were without the monthly fee so you would do:

50.50-18=32.50

so 32.50 is the price of all the phone calls

then you divide 32.50 by 0.05 which equals to 650

meaning that n=650

hope I helped!

5 0
3 years ago
Can anyone please help me solve this question!!!
sineoko [7]

Answer:

KL = 4

Step-by-step explanation:

JL = 4

3 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Clarge%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%
SVETLANKA909090 [29]

\huge \boxed{\mathbb{QUESTION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)

The remaining elements of the product matrix are found in the same way.

\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)

Simplify each element by multiplying the individual terms.

\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)

Now, sum each element of the matrix.

\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}

7 0
3 years ago
If f(x) = 3x-5 and g(x) = 4x^2 + 2x -8 evaluate f(x) x g(x)
Vikentia [17]

(3x-5)*(4x^2+2x-8)

=12x^3−14x^2−34x+40


3 0
3 years ago
Is 23 prime or composite ?
dalvyx [7]
It is prime, it has no factors besides 1,23
3 0
3 years ago
Read 2 more answers
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