Answer:
a). Y = y0e^-k(t)
b) Y = 19.4 Unit mass
Step-by-step explanation:
Y = y0e^-k(t)
Where y is amount present at the time
Y0 is initial amount present at t = 0
Y0 = 58.7
Half life = 5 hours
At half life , y = 58.7/2
At half life , y = 29.35
K = decaying constant.
Let's look fithe value of k
Y = y0e^-k(t)
29.35 = 58.7e^-k(5)
29.35/58.7 = e^-k(5)
0.5 = e^-k(5)
In 0.5 = -k(5)
-0.69314718 = -k(5)
0.138629436 = k
The value present in 8 hours will be
Y = y0e^-k(t)
Y = 58.7e-0.138629436(8)
Y = 58.7e-1.109035488
Y = 58.7(0.329876978)
Y= 19.36377861
To the nearest tenth
Y = 19.4 unit of mass
Answer:
The output is given as
Step-by-step explanation:
As the output is given as y(n) and the input is given as x(n) thus the
equation of the output is given as
Here the impulse response h(n) is given as
So the output is given as
By distributive property,
Now by the convolution properties with delta
so
So the output is given as
Answer:
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Step-by-step explanation:
Answer with Step-by-step explanation:
Since we have given
n = 16
a) Describe the sampling distribution for the sample mean.
The sampling distribution of the mean is the population's mean from where the items are sampled.
It would be
b. What is the standard error?
c. For 90% confidence, what is the margin of error?
Degrees of freedom = v - 1= 16-1 = 15
Using t-table, 1.753 cuts 5% level of significance in each tail.
so, Margin of error would be
d. Based on the sample results, create the 90% confidence interval and interpret.
So, 90% confidence interval would be
(103.86,116.136)