Answer: D. 74.91
One pound is 2.205 kilograms, so we can find how many kilograms are in 165 lbs by dividing 165 by 2.02.
The answer is approximately 74.91
No (unless you're talking about a whole number), for example...
0.2^2 = 0.04
Answer:
The Probability of the pebble landing on red = 1/3
The Probability of the pebble landing on blue = 1/6
The Probability of the pebble landing on white = 1/2
Total sum of probability = 1/3 + 1/6 + 1/2 = 1
The probability that the first pebble lands on blue and second pebble lands on red is:
1/6 * 1/3
= 1/18
Step-by-step explanation:
Red, Red = 1/3 * 1/3
Red, Blue = 1/3 * 1/6
Red, white = 1/3 * 1/2
Blue, Blue = 1/6 * 1/6
Blue, Red = 1/6 * 1/3
Blue, white = 1/6 * 1/2
White, White = 1/2 * 1/2
White, Red = 1/2 * 1/3
White, Blue = 1/2 * 1/6
Answer:
Step-by-step explanation:
<u>21. 1st scenario</u>
- B = fy + wz
- 100,000 = 15,000*4 + 500w
- 500w = 100,000 - 60,000
- 500w = 40,000
- w = 40,000/500
- w = 80
80 tons of wheat
<u>22. 2nd scenario</u>
- B = fy + wz
- 50,000 = 2y + 200*50
- 50,000 = 2y + 10,000
- 2y = 50,000 - 10,000
- 2y = 40,000
- y = 20,000
Fighter plane's price is $20,000
Let's call the two numbers
and
.
Given these variables, we can say:
, based on the first sentence in the problem.
Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

To solve, we can simply substitute
in for
in the second equation and solve.


- Get terms on the left side to a common denominator for easier addition


- Cross multiplication (
)


- Subtract
from both sides of the equation

- Factor left side of the equation

Now, notice that we have found two solutions, but the problem is only asking for one. This <em>likely </em>means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,
,
Since
is not positive in this case,
is not a solution.
Thus,
is our only solution. In this case,
,
which means that the smaller number is 14 and the larger number is 28.