Answer:
The correct answer is:
44% (B.)
Step-by-step explanation:
number of times flipped = 50 times
number of times red landed face-up = 22 times
∴ frequency of chip landing with red side face up = (number of times red landed face-up) ÷ (number of times flipped) × 100
= (22 ÷ 50) × 100 = 44%
Answer:
$120.44
Step-by-step explanation:
100.37 + 20% (20.07)=$120.44
Answer:
A
Step-by-step explanation:
10^-3 will make it negative and then 10^3 because it is less then 10^7
Hello there!
<u>F</u><u>o</u><u>r</u><u>m</u><u>u</u><u>l</u><u>a</u><u>s</u>

First, we have to find cos(theta). We know that the theta lies in Q3 which is negative for both sin and cos. To find cos, you can use the identity or Pythagoras.
<u>R</u><u>e</u><u>c</u><u>a</u><u>l</u><u>l</u>

Finding cos, I will use Pythagoras.
<u>R</u><u>e</u><u>c</u><u>a</u><u>l</u><u>l</u><u> </u><u>P</u><u>y</u><u>t</u><u>h</u><u>a</u><u>g</u><u>o</u><u>r</u><u>e</u><u>a</u><u>n</u>
- Define a or b = adjacent or opposite
- Define c = hypotenuse

From sin = -1/6
- opposite = 1
- hypotenuse = 6
- adjacent = x

b is our adjacent. Hence,

Since theta is in Q3, cos < 0 or in negative.

<u>F</u><u>i</u><u>n</u><u>d</u><u>i</u><u>n</u><u>g</u><u> </u><u>s</u><u>i</u><u>n</u><u>(</u><u>2</u><u>t</u><u>h</u><u>e</u><u>t</u><u>a</u><u>)</u>

Hence,

<u>R</u><u>e</u><u>c</u><u>a</u><u>l</u><u>l</u>
- Negative × Negative = Positive
<u>F</u><u>i</u><u>n</u><u>d</u><u>i</u><u>n</u><u>g</u><u> </u><u>c</u><u>o</u><u>s</u><u>(</u><u>2</u><u>t</u><u>h</u><u>e</u><u>t</u><u>a</u><u>)</u>
The cos of 2theta has multiple formulas. All of them are same and also identity to each others.
I will be using the first one as it only involves cos, not sin.

Therefore,

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
- sin(2theta) = √35/18
- cos(2theta) = 17/18
The last choice.
Let me know if you have any questions!
D)

This is because it contains more than 2 algebraic terms