The sum of y and 20 is less than 37

Mathematics

1 answer:

nata0808 [166]3 years ago

8 0

Y + 20 < 37

y < 17 the answer to the question

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What is P(2 or divisor of 9)?<br> please explain!

Dmitrij [34]

The probability that a divisor of 2 or 9 is chosen when the spinner is spun once is 2/3.

<h3>What is the probability?</h3>

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

P(2 or divisor of 9) = section that has a divisor of 2/ total number of sections + section that has a divisor of 2/ total number of sections

1/3 + 1/ 3 = 2/3

To learn more about probability, please check: brainly.com/question/13234031

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3 0

2 years ago

A circle is centered at J(3, 3) and has a radius of 12.

stealth61 [152]

Answer:

$(-6,\, -5)$ is outside the circle of radius of $12$ centered at $(3,\, 3)$ .

Step-by-step explanation:

Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$ .

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$ , then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$ .

Point $F$ would be inside this circle if $d(J,\, F) < r$ . (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$ . (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$ . (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$ :

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$ .

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$ , is smaller than $d(J,\, F)$ . Therefore, point $F$ would be outside this circle.