Ask question

Ask question

All categories

3 years ago

11

Joseph buys a bag of 60 pieces of taffy. In the bag, there are 6 cinnamon pieces, 18 raspberry pieces, 12 key lime pieces, and t

he rest are candy apple pieces. If he randomly draws one piece of taffy, what is the probability that it is a piece of key lime taffy?
A. 0.1
B. 0.4
C. 0.02
D. 0.2

1 answer:

Gala2k [10]3 years ago

7 0

To find the probability of Joseph picking a key lime piece of taffy you will create a fraction with the number of possible outcomes over the total possible outcomes.

<u># of key lime taffies</u> <u>12</u>
total # of taffies = 60

12/60 = 1/5

1/5 = 0.2

Joseph has a 0.2 probability of picking a key lime taffy.

You might be interested in

Translate point P(3,3) left 1 unit and up 4 units. What are the coordinates of the image P’?

ANTONII [103]

The coordinates would be (2,7) you just have to move 1 left 4 up

3 0

3 years ago

Are the functions linear or nonlinear?

inna [77]

Linear. They all follow the format y=mx+c wether or not they have been rearranged.

5 0

3 years ago

Read 2 more answers

Change Y = 3X to the standard form of the equation of a line.

ehidna [41]

Answer:

-3x+y=0

Step-by-step explanation:

Standard form= Ax+Bx=C

Add -3x both sides, answer -3x+1y=0

3 0

3 years ago

Question 9 of 10

allochka39001 [22]

Answer:

I think it's 126cm not sure

7 0

2 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$ .

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$ .

Solve this equation for $u$ :

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$ .

$u^{2} + 1 = 2 u^{2}$ .

$u^{2} = 1$ .

Given that $0 \le u \le 1$ , $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$ ,

$x = k \pi$ , where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$ ,

$0 \le k$ .

$k = 0$ or $k = 1$ . Accordingly,

$x = 0$ or $x = \pi$ .

8 0

2 years ago

Read 2 more answers