Mathematics, 28.10.2019 14:31 allyssaharrisooy50au

Sherwin throws a dart at this square-shaped target:

part a: is the probability of hitting the black circle inside the target closer to 0 or 1? why?

part b: is the probability of hitting the white portion of the target closer to 0 or 1? why?

Sherwin throws a dart at this square-shaped target: part a: is the probability of hitt

Answers: 2

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Answer from: Jacques1b

PartA. P_c = 0.0267

PartB. P_b = 0.9733

Step-by-step explanation:

To answer this question we must find the area of the square and the area of the circle.

The area of the circle is:

$A_c = \pi r^2$

Where r is the radius, and: r = diameter/2 and the diameter = 4

So:

r = 2

Therefore the area A_c of the circle is:

$A_c = \pi(2)^2\\\\A_c = 12.566$

Then, the area of the square is:

A_b = 22^2 -A_c

A_b = 22^2 - 12.566

A_b = 471.434

Part A

The probability P_c that the dart falls within the black circle is:

$P_c = \frac{A_c}{A_b}$

$P_c = \frac{12.566}{471.434}$

P_c = 0.0267

The probability is much closer to 0 than to 1

Part B.

The probability P_b that the dart falls within the white area is:

$P_b = 1-P_c\\\\P_b = 1-0.0267\\\\P_b = 0.9733$

The probability is much closer to 1 than to 0. This is because the area of the black circle is much smaller than the white area, therefore it is more difficult for the dart to hit the black circle.

Answer from: Quest

the correct answer is b. negative exponential growth.

step-by-step explanation:

you can tell it's an exponential function because it decreases rapidly then levels out and never touches the x-axis.

it is negative because the y-axis decreases as the x-axis increases.

therefore it is a negative exponential growth graph.

hope i !

Categorize the graph as linear increasing, linear decreasing, exponential growth, or negative expone

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

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Sherwin throws a dart at this square-shaped target:

part a: is the probability of hitt...

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