Terbium-160 has a half-life of about 72 days. After 396 days, about how many milligrams of a 220 mg sample will remain?
2 answers:
Answer:
Option A = 5 mg
Step-by-step explanation:
Given : Terbium-160 has a half-life of about 72 days.
To find : After 396 days, about how many milligrams of a 220 mg sample will remain?
Solution :
We have given the Terbium-160 has a half-life of about 72 days.
We can represent the situation with an exponential function,
Where,
is the amount at any time t,
is the original amount,
n=72 is the half-life
t=365 number of days
Substituting all the values,
Approximately 4.84=5 mg
Therefore, Option A is correct.
After 396 days, there will only be 5 mg of Terbium-160.
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Answer:
a
The population parameter of interest is the true proportion of Greek who are suffering
While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%
b
The condition is met
c
The 95% confidence interval is
d
If the confidence level is increased which will in turn reduce the level of significance but increase the critical value() and this will increase the margin of error( deduced from the formula for margin of error i.e
) which will make the confidence interval wider
e
Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower
Step-by-step explanation:
From the question we are told that
The sample size is n = 1000
The population proportion is
Considering question a
The population parameter of interest is the true proportion of Greek who are suffering
While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%
Considering question b
The condition for constructing a confidence interval is
So
Hence the condition is met
Considering question c
Given that the confidence level is 95% then the level of significance is mathematically evaluated as
Next we obtain the critical value of from the normal distribution table, the value is
Generally the margin of error is mathematically represented as
substituting values
The 95% confidence interval is mathematically represented as
substituting values
substituting values
considering d
If the confidence level is increased which will in turn reduce the level of significance but increase the critical value() and this will increase the margin of error( deduced from the formula for margin of error i.e
) which will make the confidence interval wider
considering e
Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower
The distance between a point 
on the given plane and the point (0, 2, 4) is
but since
and 
share critical points, we can instead consider the problem of optimizing subject to 
.
The Lagrangian is
with partial derivatives (set equal to 0)
Solve for
:
which gives the critical point
We can confirm that this is a minimum by checking the Hessian matrix of:
is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.
At this point, we get a distance from (0, 2, 4) of
but since
The Lagrangian is
with partial derivatives (set equal to 0)
Solve for
which gives the critical point
We can confirm that this is a minimum by checking the Hessian matrix of
At this point, we get a distance from (0, 2, 4) of
The value of x is 44°.
Step-by-step explanation:
Step 1:
The sum of the angles in a triangle is equal to 180°.
The sum of the angles along a straight line is 180°.
The two given angles of the triangle are 14° and 30°.
Assume the third angle is a.
The sum of 14°, 30° and a is 180°.
So the third angle of the triangle is 136°.
Step 2:
The sum of the angles along a straight line is 180°.
The sum of the angle of the triangle, 136° and x is 180°.
So x measures 44°.