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3 years ago

8

If the average of 8 numbers is 12, what is the sum of the 8 numbers?

2 answers:

QveST [7]3 years ago

8 0

The average of a set of numbers is found by dividing the sum of the numbers by the amount of numbers in the set.

If the average of 8 numbers is 12, that means that the sum of the numbers divided by 8 is equal to 12.

To find the sum, we need to multiply 12 by 8:

sum/8 = 12

(sum/8)*8 = 12*8

sum = 96

The answer is 96.

Sholpan [36]3 years ago

4 0

Either I don't understand the question fully or this has a whole lot of answers.
For example, if all the numbers were 12 there would be an average of 12 and would equal 96 as the sum

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Step-by-step explanation:

$n-63=126\\n=126+63\\n=189$

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2 years ago

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3 years ago

Find the 12th term of the geometric sequence 5, -25, 125, ...5,−25,125,...

katovenus [111]

Answer:

$a_{12}=-244140625$

Step-by-step explanation:

Considering the geometric sequence

$5,-25,\:125,\:...$

$a_1=5$

As the common ratio ' $r$ ' between consecutive terms is constant.

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$r=\frac{-25}{5}=-5$

$r=\frac{125}{-25}=-5$

The general term of a geometric sequence is given by the formula:

$a_n=a_1\cdot \:r^{n-1}$

where $a_1$ is the initial term and $r$ the common ratio.

Putting $n = 12$ , $r = -5$ and $a_1=5$ in the general term of a geometric sequence to determine the 12th term of the sequence.

$a_n=a_1\cdot \:r^{n-1}$

$a_n=5\left(-5\right)^{n-1}$

$a_{12}=5\left(-5\right)^{12-1}$

$=5\left(-5^{11}\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=-5\cdot \:5^{11}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$=-5^{1+11}$ ∵ $5\cdot \:5^{11}=\:5^{1+11}$

$=-244140625$

Therefore,

$a_{12}=-244140625$

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