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TiliK225 [7]
3 years ago
12

In one of his orchards, a farmer harvested 270 apples but realized that 80 of them were unfit for consumption because they were

either overripe or ruined by insects. What percentage of apples were actually good for consumption?
Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0

Answer: 70.37\%

Step-by-step explanation:

Let be "x" the percentage of apples  that were actually good for consumption.

You know that total number of apples the farmer harvested was 270 apples.  

Then, you can identify that 270 apples represents the 100\%

You know that 80 of them were unfit for consumption, so the number of apples  that were good for consumption was:

270\ apples-80\ apples=190\ apples

Then, you can set up the  following proportion in order to find the percentage of apples  that were actually good for consumption:

\frac{100}{270}=\frac{x}{190}

Finally, you must solve for "x" in order to find its value. This is:

(190)(\frac{100}{270})=x\\\\x=70.37\%

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Step-by-step explanation:

1. One ton is 2,000 pounds. So you would have to multiply 5.5 x 2,000.

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Answer:

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Step-by-step explanation:

a) Geometrically speaking, the average rate of change of y with respect to x over the interval by definition of secant line:

r = \frac{y(b) -y(a)}{b-a} (1)

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y(a), y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that y = 3\cdot x^{2}, a = 3 and b = 6, then the average rate of change of y with respect to x over the interval is:

r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}

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The average rate of change of y with respect to x over the interval [3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:

y' =  \lim_{h \to 0}\frac{y(x+h)-y(x)}{h} (2)

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h - Change rate.

y(x), y(x+h) - Function evaluated at x and x+h.

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y' = 3\cdot  \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}

y' = 6\cdot  \lim_{h \to 0} x +3\cdot  \lim_{h \to 0} h

y' = 6\cdot x

y' = 6\cdot (3)

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