The weight of a radioactive isotope was 96 grams at the start of an experiment. After one hour, the weight of the isotope was ha
lf of its initial weight. After two hours, the weight of the isotope was half of its weight the previous hour. If this pattern continues, which of the following graphs represents the weight of the radioactive isotope over time?
2 answers:
The starting weight of the radioactive isotope = 96 grams
Weight after one hour is half of the starting weight. So the weight of the radioactive isotope after 1 hour = 48 grams
After 2 hours the weight is half as compared to the weight after previous hour. So weight after 2 hours = 24 grams.
This means, after every hour the weight is being halved. The half life of radioactive isotope is one hour.
Since after every hour, the weight is being halved, the weight of the isotope can be modeled by an exponential equation.
So,
Initial Weight = W₁ = 96
Change factor = 1/2 = 0.5
The general equation of the sequence will be:
Here t represents the number of hours. Using various values of t we can find the weight of the radioactive isotope at that time.
We can plot the sequence using the above equation. The graph is attached below.
Weight after one hour is half of the starting weight. So the weight of the radioactive isotope after 1 hour = 48 grams
After 2 hours the weight is half as compared to the weight after previous hour. So weight after 2 hours = 24 grams.
This means, after every hour the weight is being halved. The half life of radioactive isotope is one hour.
Since after every hour, the weight is being halved, the weight of the isotope can be modeled by an exponential equation.
So,
Initial Weight = W₁ = 96
Change factor = 1/2 = 0.5
The general equation of the sequence will be:
Here t represents the number of hours. Using various values of t we can find the weight of the radioactive isotope at that time.
We can plot the sequence using the above equation. The graph is attached below.
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Let and
represents the proportions of the seeds which germinate among the seeds planted in the soil containing
and
mushroom compost by weight respectively.
To test the null hypothesis against the alternate hypothesis
.
Let denotes the respective sample proportions and the
represents the sample size respectively.
The test statistic can be written as :
which under follows the standard normal distribution.
We reject at
level of significance, if the P-value
or if
Now, the value of the test statistics = -1.368928
The critical value =
P-value =
= 0.171335
Since the p-value > 0.05 and , so we fail to reject
at
level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the of the seeds that
with the percent of the
in the soil.
Step-by-step explanation:
For this parallelogram to exist least value of x should be 4.
Because if x = 3
(x - 3)= 3 - 3 =0
Thus width of parallelogram would become zero as a result area of parallelogram would become zero i. e. Parallelogram won't exist.