Step-by-step explanation:
Here's the first one for ya. I can't tell what the first denominator of the second one is [CosB, CscB, or SecB?], so unless you get back to me on that, then I can't help you with it.
Answer:
The child has 6 dimes and 5 quarters.
Step-by-step explanation:
Let q and d represent the number of quarters and of dimes respectively.
Then q + d = 11 (Equation A), and ($0.25/quarter)q + ($0.10/dime)d = $1.85 (Equation B).
Multiply the 2nd equation by 100 to remove the decimal fractions:
25q + 10d = 185 (Equation C)
Now multiply the 1st equation by -10 to obtain -10q - 10d = -110 (Equation D), and combine this result with Equation C:
-10q - 10d = -110
25q + 10d = 185
--------------------------
15q = 75, and so q = 75/15 = 5.
According to Equation A, q + d = 11. Replacing q with 5, we get:
5 + d = 11, and so d = 6.
The child has 6 dimes and 5 quarters.
Answer:
See below ↓↓↓
Step-by-step explanation:
<u>Probability of getting heads</u>
- Outcome (head) = 1
- Total outcomes = 2
- P (head) = <u>1/2</u>
<u></u>
<u>Probability of getting an odd number</u>
- Outcome (odd) = {1, 3, 5} = 3
- Total outcomes = 6
- P (odd) = 3/6 = <u>1/2</u>
<u></u>
The outcomes are equally likely because the probabiity of each outcome is the same.
