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3 years ago

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in a group of 120 people,90 of them are 30 years old or older. Everyone else in the group is younger than 20 years old. If a per

son is selected at random, what is the probability the person is less than 20 years old

1 answer:

Ksju [112]3 years ago

3 0

36% chance of the person being less than 20 years okd

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Write an equation in standard form passing through the points (-2, 0) and (-3, -1). I know the answer to the question, but I do

g100num [7]

Answer:

$y - x = 2$

Step-by-step explanation:

Given

Points (-2,0) and (-3,-1)

Required

Equation of Line

The first step is to determine the slope of the line; using the following formula

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$Where\ (x_1,y_1) = (-2,0)\ and\ (x_2,y_2) = (-3,-1)$

$m = \frac{y_2 - y_1}{x_2 - x_1}$ becomes

$m = \frac{-1 -0}{-3 - (-2)}$

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$m = \frac{-1}{-1}$

$m = 1$

The next is to determine the equation of the line using any of the points

The formula is as thus;

$m = \frac{y - y_1}{x - x_1}\ or\ m = \frac{y - y_2}{x - x_2}$

Using $m = \frac{y - y_1}{x - x_1}$

$Where\ (x_1,y_1) = (-2,0)\ and\ m =1$

$1 = \frac{y-0}{x - (-2)}$

$1 = \frac{y-0}{x +2}$

$1 = \frac{y}{x +2}$

Multiply both sides by x + 2

$(x+2)*1 = \frac{y}{x +2} * (x+2)$

$(x+2)*1 = y$

$x+2 = y$

$y = x + 2$

Subtract x from both sides

$y - x = x - x +2$

$y - x = 2$

Hence, the equation of the line in standard form is $y - x = 2$

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The first term of a geometric sequence is equal to a and the common ratio of the sequence is r.

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Answer: (a) {a, ar, ar², ar³, ar⁴, ar⁵...}, (b) arⁿ⁻¹

For part (a), the question gives us the first term a, and then asks us to apply the common ratio r six times.

In order for ar = a, the nth term of r will have to equal 0 (this implies that n is an exponent; thus giving us the first term a, as r = 1).

Since we use this method on the first term, we must use it for the next five, in which r gains an additional exponent for every consecutive value (nth term) thereafter.

Ultimately getting: {a, ar, ar², ar³, ar⁴, ar⁵...}

For part (b), we first have to understand that the sequence does not start at 0, but at 1 for n. In order for ar = a, with n = 1, there needs to be subtraction of -1 within the exponent. So that arⁿ⁻¹

If we check and apply this, we can see that:

{ar¹⁻¹, ar²⁻¹, ar³⁻¹, ar⁴⁻¹, ar⁵⁻¹, ar⁶⁻¹...} = {a, ar, ar², ar³, ar⁴, ar⁵...} = arⁿ⁻¹ = Tn

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