Which is one of the transformations applied to the graph of f(x) = x2 to change it into the graph of g(x) = 4x2 + 24x + 30?
2 answers:
The graph is shifted left 3 units.
We know that the graph of f(x) is NOT widened, because for a number c greater than 1, cf(x) stretches the graph vertically, not horizontally.
We know that the graph of f(x) is NOT shifted up 30 units. It is tempting to say this because we do see +30 as the last term. But, we have to calculate the vertex of g(x). If we find the point
, that is, by taking doing 
, 
, we see that the vertex is six units down from f(x), not 30 units up!
We know that the graph of f(x) is NOT reflected over the x-axis, because there would need to a negative coefficient on the
-term, and there isn't.
We know that the graph of f(x) IS shifted 3 units to the left. When we calculated the vertex earlier, we saw that it is
, which indeed corresponds to a shift 3 units to the left from the origin.
We know that the graph of f(x) is NOT widened, because for a number c greater than 1, cf(x) stretches the graph vertically, not horizontally.
We know that the graph of f(x) is NOT shifted up 30 units. It is tempting to say this because we do see +30 as the last term. But, we have to calculate the vertex of g(x). If we find the point
We know that the graph of f(x) is NOT reflected over the x-axis, because there would need to a negative coefficient on the
We know that the graph of f(x) IS shifted 3 units to the left. When we calculated the vertex earlier, we saw that it is
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Answer:
a)0.976
b)0.00926
c)0.2402
d)0.35
Step-by-step explanation:
Let be an item passed by inspector i
Let Y be the event that there is a fault in an item
The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1
If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.
So,
If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.
So,
If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e.
So,
If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e.
So,
a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )
P(found by atleast one inspector | It has flaw )=
P(found by atleast one inspector | It has flaw )=
P(found by atleast one inspector | It has flaw )=0.976
Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976
b)
C)P( two inspectors draw different conclusions on the same item)=
P( two inspectors draw different conclusions on the same item)=0.2402
D)