The answer is D. The grand total for 12th graders is 68. Divide the 28 students in 12th grade who prefer math by the grand total to get you a rounded answer of 0.412.
Answer:
a) The half life of the substance is 22.76 years.
b) 5.34 years for the sample to decay to 85% of its original amount
Step-by-step explanation:
The amount of the radioactive substance after t years is modeled by the following equation:

In which P(0) is the initial amount and r is the decay rate.
A sample of a radioactive substance decayed to 97% of its original amount after a year.
This means that:

Then



So

(a) What is the half-life of the substance?
This is t for which P(t) = 0.5P(0). So







The half life of the substance is 22.76 years.
(b) How long would it take the sample to decay to 85% of its original amount?
This is t for which P(t) = 0.85P(0). So







5.34 years for the sample to decay to 85% of its original amount
Answer:
A = 90.25 pi in^2
Step-by-step explanation:
The circumference is given by
C = 2*pi*r
19pi = 2*pi*r
Divide each side by 2 pi
19pi / 2 pi = 2*pi*r/ 2pi
9.5 = r
The area is given by
A = pi r^2
A = pi ( 9.5)^2
A = 90.25 pi in^2
Answer:
mid point=(x1 + x2)/2 (y1 + y2)/2
(-5+5)/2 and (-7+4)/2
0 and 1.5
midpoint=(0,1.5)
For this case what you should do is use the given expression and add 14.
We have then:
Original function:
f (p) = p + 11.
New function:
g (p) = f (p) +14
Substituting we have:
g (p) = (p + 11) +14
g (p) = p + 25
Answer:
An expression that represents the total mumber of pots Noah made one weekend if he made 14 more than his usual number is:
g (p) = p + 25