62.5 mg sample will remain after 240 days
Step-by-step explanation:
Given
Half-life = T = 60 days
The formula for calculating the quantity after n half lives is given by:

Here
N is the final amount
N_0 is the initial amount
n is the number of half lives passed
The number of half lives are calculated by dividing the time for which the remaining quantity has to be found by half life
The quantity has to be calculated for 240 days so,

Given

Putting the values in the formula

Hence,
62.5 mg sample will remain after 240 days
Keywords: Half-life, sample
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Answer:
4
Step-by-step explanation:
The diagram shows that the one triangle can be divided into two equal right triangles. Because of this, you can use the Pythagorean Theorem to solve this problem. a and b are the two sides of the angles, and c is the hypotenuse.
The given lengths are 5 as the hypotenuse and 3 as one length. (You have 3 as a given length because the two triangles have a length of 6 on one side. 6/2 = 3)
a² + b² = c²
a² = c² - b²
a² = 5² - 3²
a² = 25² - 9²
a² = 16
a = 4
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
Answer:
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