Ab=81, a=81/b
s=a+b using a from above in this we get:
s(b)=81/b +b
s(b)=(81+b^2)/b
ds/db=(2b*b-81-b^2)/b^2
ds/db=(2b^2-81-b^2)/b^2
ds/db=(b^2-81)/b^2
d2s/db2=(2b^3-2b^3+81)/b^4
d2s/db2=81/b^4 since b is positive we know that the acceleration is positive so that when ds/db=0 it is a minimum for s(b)
ds/db=0 only when b^2-81=0, b^2=81, b=9
The two positive numbers are 9 and 9.
Answer:
0.3902
Step-by-step explanation:
The question meets the criteria for the calculation of the confidence interval of a one sample proportion ;
Sample size, n = 27
Number of students with faster reaction time = 14
Phat = x / n = 14 / 27 = 0.6296
The confidence interval is calculated thus :
C. I = Phat ± Zcritical * √[p(1 - p)/n]
Zcritical at 99% = 2.576
C. I = 0.6296 ± 2.576 * √[0.6296(0.3704)/27]
C.I = 0.6296 ± 0.2394042
The lower boundary of C.I = 0.6296 - 0.2394042
Lower boundary = 0.3902
Answer:
C is the correct answer.
Step-by-step explanation:
Answer:
1 and 1/3
Step-by-step explanation:
3a² -4a +1 = 0
3a²- 3a - a+ 1= 0
3a(a-1) - (a-1)= 0
(a-1)(3a -1)= 0
a-1= 0 ⇒ a= 1
3a - 1= 0 ⇒ 3a= 1 ⇒ a= 1/3
Hello,
Answer:
1) given equation
2) distributive property
3) combing like terms
4)addition or subtraction property of equality
5) multiplication or division property of equality
Explanation:
Hope this helps!