Hello!
Here are some rules to determine the number of significant figures.
- Numbers that are not zero are significant (45 - all are sigfigs)
- Zeros between non-zero digits are significant (3006 → all are sigfigs)
- Trailing zeros are not significant (0.067 → the first two zeros are not sigfigs)
- Trailing zeros after a decimal point are always significant (1.000 → all are sigfigs)
- Trailing zeros in a whole number are not significant (7800 → the last two zeros are not sigfigs)
- In scientific notation, the exponential digits are not significant, known as place holders (6.02 x 10² → 10² is not a sigfig)
Now, let's find the number of significant figures in each given number.
A). 296.54
Since these digits are all <em>non-zero</em>, there are 5 significant figures.
B). 5003.1
Since the two <em>zeros are between non-zero digits</em>, they are significant figures. Thus, there are 5 significant figures.
C). 360.01
Again, the two zeros are between non-zero digits. There are 5 significant figures.
D). 18.3
All of these digits are non-zero, hence, there are 3 significant figures.
Therefore, expression D has the fewest number of significant figures being 3.
For this case we have to, by defining properties of powers and roots the following is fulfilled:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
We must rewrite the following expression:
![\sqrt [3] {8 ^ {\frac {1} {4} x}}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D)
Applying the property listed we have:
![\sqrt [3] {8 ^ {\frac {1} {4} x}} = 8 ^ {\frac{\frac {1} {4} x} {3} }= 8 ^ {\frac {1} {4 * 3} x} = 8 ^ {\frac {1} {12} x}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D%20%3D%208%20%5E%20%7B%5Cfrac%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%20%7B3%7D%20%7D%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%20%2A%203%7D%20x%7D%20%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D)
Using the property again we have to:
![8 ^ {\frac {1} {12} x} = \sqrt [12] {8 ^ x}](https://tex.z-dn.net/?f=8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D%20%3D%20%5Csqrt%20%5B12%5D%20%7B8%20%5E%20x%7D)
Thus, the correct option is option C
Answer:
Option C
There is a fifty percent chance of the coin landing on "heads" each time it is flipped.
However, flipping a coin 20 times virtually guarantees that it will land on "heads" at least once in that twenty times. <span>(99.9999046325684 percent chance)
</span>
You can see this by considering two coin flips. Here are the possibilities:
Heads, heads.
Heads, tails.
Tails, tails.
Tails, heads.
You will note in the tossing of the coin twice that while each flip is
fifty/fifty, that for the two flip series, there are three ways that it
has heads come up at least once, and only one way in which heads does
not come up. In other words, while it is a fifty percent chance
for heads each time, it is a seventy five percent chance of seeing it
be heads once if you are flipping twice. If you wish to know
the odds of it not being heads in a twenty time flip, you would multiply
.5 times .5 times .5...twenty times total. Or .5 to the twentieth
power. That works out to a 99.9999046325684 percent chance of
it coming up heads at least once in the twenty times of it being
flipped.
Answer: 4,988 porque 16% de 4,300 es 688 asi que nomas lo aggregas a lo que costaba y agaras el resultado de 4,988
Step-by-step explanation:
The best scale would probably be going by 10s. You should start at around 80 and go up to 160.
