Answer:
We have the next relation:
A = (b*d)/c
because we have direct variation with b and d, but inversely variation with c.
Now, if we have 3d instead of d, we have:
A' = (b*(3d))/c
now, we want A' = A. If b,c, and d are the same in both equations, we have that:
3bd/c = b*d/c
this will only be true if b or/and d are equal to 0.
If d remains unchanged, and we can play with the other two variables we have:
3b'd/c' = bd/c
3b'/c' = b/c
from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.
Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c
Answer:
7. 28.8722
8. 21.7093
Step-by-step explanation:
458.29 x .063=28.8722
21.7093 x .0775= 21.7093
So,
The probability of the man having diabetes is 0.6 or 60%. Because we are figuring the probability BEFORE the test is taken that he has the disease, we can disregard the test and its accuracy rate. That rate is 60%, the probability of him having diabetes.
The correct option is D.
Answer:
Step-by-step explanation:
as given in question that a > 0 so
if we put a=1
we get g(x) = f(x)
now put a =2
we get
g(x) = 2 f(x)
here we can see that g(x) would always be greater than or equals to f(x)
so we can say that the graph of g(x) will never be narrower than graph of g(x)
