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Olenka [21]
2 years ago
13

A small dairy farm writes a business plan. It includes a graph showing the estimated amount of feed used each day in thousands o

f pounds, y, compared to the number of years since the farm began operating, x.
What does the y-intercept represent?

the estimated amount of years the farm can afford feed
the maximum feed per day estimated for the life of the farm
the estimated amount of feed per day for the start of the farm
the maximum years the farm needs the originally estimated feed per day
Mathematics
1 answer:
zavuch27 [327]2 years ago
3 0

The y intercept means , the estimated amount of feed per day for the start of the farm , Option C is the right answer.

<h3>What is a function ?</h3>

A function is a law that relates two variables , independent and dependent variable.

It is given that

On y axis , amount of feed used each day in thousands of pounds,

On x axis,  number of years since the farm began operating

The y intercept means x = 0 , which means at the starting of the farm , the feed needed per day.

Therefore Option C is the right answer , the estimated amount of feed per day for the start of the farm.

To know more about Function

brainly.com/question/12431044

#SPJ1

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An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ
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a) P=0.558

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a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558

b) The approximate probability that at least 9 carry the gene is:

P(x\geq9)=1-P(x\leq 8)\\\\\\

P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

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