Answer:
2.25 %
Explanation:
65-95-99.7 is a rule to remember the precentages that lies around the mean.
at the range of mean (
) plus or minus one standard deviation (
), ![P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%](https://tex.z-dn.net/?f=P%28%5B%5Cmu-%5Csigma%20%5Cleq%20X%20%5Cleq%20%5Cmu%2B%5Csigma%5D%29%5Capprox%2068.3%25)
at the range of mean plus or minus two standard deviation, ![P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%](https://tex.z-dn.net/?f=P%28%5B%5Cmu%20-2%5Csigma%20%5Cleq%20X%20%5Cleq%20%5Cmu%2B2%5Csigma%5D%29%5Capprox%2095.5%25)
at the range of mean plus or minus three standard deviation, ![P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%](https://tex.z-dn.net/?f=P%28%5B%5Cmu%20-%203%5Csigma%5Cleq%20X%20%5Cleq%20%5Cmu%2B3%5Csigma%5D%29%5Capprox%2099.7%25)
So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (
)
So we know that the 95.5% is between 
and 
, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.
So 
Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.
The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
Answer:
Kinetic energy is given by:
K.E. = 0.5 m v²
Susan has mass, m = 25 kg
Velocity with which Susan moves is, v = 10 m/s
Hannah has mass, m' = 30 kg
Velocity with which Hannah moves is, v' = 8.5 m/s
<u>Kinetic energy of Susan:</u>
0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J
<u>Kinetic energy of Hannah:</u>
0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J
Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.
Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.
Answer:
binding energy is 99771 J/mol
Exlanation:
given data
threshold frequency = 2.50 × 
Hz
solution
we get here binding energy using threshold frequency of the metal that is express as
..................1
here E is the energy of electron per atom 
and h is plank constant i.e. 
and x is binding energy
and here N is the Avogadro constant = 
so E will 
E = 
so put value in equation 1 we get
= 2.50 × ×
solve it we get
x = 99770.99
so binding energy is 99771 J/mol
