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aliya0001 [1]
1 year ago
12

The size of a population of fish in a pond ismodeled by the function P, where P(t) givesthe number of fish and t gives the numbe

ryears after the first year of introduction of thefish to the pond for 0 < t < 10. The graphof the function P and the line tangent to P att= 4 are shown above. Which of thefollowing gives the best estimate for theinstantaneous rate of change of P at t=4?

Mathematics
1 answer:
Goryan [66]1 year ago
7 0
<h2>Answer:</h2>

The slope of the line joining (3.9, P(3.9)) and (4.1, P(4.1)). Option D is correct.

<h2>Explanations:</h2>

Instantaneous rate of change is the measure of slope of a curve at a given instant.

From the given diagram, we can see that there is a line drawn tangential to the curve. The x-axis (time) for the slope of the curve must lie within the range of the tangential line.

Sine the range of time for the tangential line is between around t = 3 and t = 5, hence the x-axis for the slope must lie within this range.

From the given option, the only coordinate points that lie within the range is (3.9, P(3.9)) and (4.1, P(4.1))

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3 years ago
Springfield nuclear energy inc bonds are currently trading at 283.30. The bonds have a face value of $1,000, a coupon rate of 2%
vampirchik [111]

<u>Answer:</u>

The yield to maturity of the bonds is 11%

<u>Explanation:</u>

Price at which the bonds is currently trading = 283.30$

Face Value = $1000

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Hence the coupon bond rate = $1000 ×2%

= 1000\times \frac{2}{100}

=$20

Years to maturity: 20 years

Formula used:  

=\frac{(C+ (\frac{(F-P)}{n}))}{(\frac{(F+P)}{2})}

Where C is the bond coupon rate

F is the face value

P is the price

N is the number of years

=\frac{(20 +(\frac{(1000-283.30)}{20})}{(\frac{(1000+283.30)}{2})}

=11%

The yield to maturity of the bonds is 11%

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3 years ago
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What is the best approximation for the perimeter of a semicircle with a dimeter of 12cm?
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