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3 years ago

What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

Mathematics

1 answer:

NeX [460]3 years ago

7 0

Answer:

letter C: 100

Step-by-step explanation:

This problem can be configured as an arithmetic progression:

The first value is 10, the ratio is also 10 (the multiples of ten are 20, 30, 40, ...) and the number of terms is 19 (10, 20, 30, ... , 180, 190 -> total of 19 numbers).

So, we can use the following formula to calculate the sum of all these numbers:

Sn = n(a1 + an) /2

where Sn is the sum, n is the number of terms, a1 is the first term and an is the last one.

The result we want is the average of the sum, so we need to divide Sn by n:

Average = Sn/n = n(a1 + an) /2n = (a1 + an) /2 = (10+190)/2 = 200/2 = 100

The correct option is letter C.

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Read 2 more answers

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$