You answered since 2009 the installed capacity of

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You Answered Since 2009, the installed capacity of solar power has been decreasing at a faster and faster rate. Since 2009, the installed capacity of solar power has been decreasing, but the rate of decrease has been declining. Since 2009, the installed capacity of solar power has been increasing at a steady rate. Correct Answer Since 2009, the installed capacity of solar power has been increasing at a faster and faster rate. S (t)>0 S′(t)>0 means the installed capacity has been increasing. S ′′ (t)>0 S″ (t)>0 means that the rate of increase has also been increasing, so installed capacity has been increasing at a faster and faster rate. Question 11 1.5 / 1.5 pts Let N=f(t) N=f(t) be the number of households (in millions) that use a 'land-line' telephone in their home, where t t is the number of years since 1995. In 1995, almost all households had a land-line telephone. By 2005 a small number of households had giving up their land-line and were relying entirely on cell-phones and internet based phone services such as Skype, but more and more households were giving up their land-line telephone service each year.
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Choose the best expression to match this statement. f (10)>0 f′(10)>0 and f ′′ (10)>0 f″(10)>0 . f (10)<0 f′(10)<0 and f ′′ (10)>0 f″(10)>0 . f (10)>0 f′(10)>0 and f ′′ (10)<0 f″(10)<0 . f (10)<0 f′(10)<0 and f ′′ (10)=0 f″(10)=0 . Correct! f (10)<0 f′(10)<0 and f ′′ (10)<0 f″(10)<0 . At t=10 t=10, (year 2005), the number of households with land-lines was decreasing so f (10)<0 f′(10)<0. More and more people giving up their land- line service means that the rate of this decrease is increasing. This means that f (t) f′(t) is becoming more negative , i.e. f (t) f′(t) is decreasing , so f(t) f(t) is concave down when t=10 t=10, so that f ′′ (10)<0 f″(10)<0. Question 12 1 / 1 pts
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If for a function f(t) f(t), the second derivative is negative ( f ′′ (t)<0 f″(t)<0), then the derivative, f (t) f′(t) is decreasing. Correct! True False True: The sign of f ′′ (t) f″(t) tells us whether f (t) f′(t) is increasing or decreasing, because f ′′ (t) f″(t) is the derivative of f (t) f′(t). So if f ′′ (t)<0 f″(t)<0, then f (t) f′(t) is decreasing, and the function f(t) f(t) is concave down. Question 13 1 / 1 pts For a function f(x) f(x), where f (2)=3 f′(2)=3, f (3)=5 f′(3)=5, and f (4)=4 f′(4)=4, it is possible for f f to be concave up on the interval (1,5) (1,5). True Correct! False False: The first derivative is both increasing and decreasing on the interval (1,5) (1,5), so the function cannot be concave up everywhere on the interval.
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Question 14 0 / 1.5 pts Let C(q) C(q) represent the cost and R(q) R(q) represent the revenue, in dollars, of producing q q items and let C (70)=26 C′(70)=26 and R (70)=24 R′(70)=24. Choose the statement that best matches this situation.
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