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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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At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a res
Ghella [55]

Answer:

a) P(F UR) = P(F) +P(R) -P(F and R) = 0.72+0.46-0.32=0.86

b) P(FUR)' = 1-P(FUR)= 1-0.86 = 0.14

Step-by-step explanation:

Let's define the following events first:

F: The event that a course has a final exam.

R: The event that a course requires a research paper

From the info provided we have that:

P(F) = 0.72, P(R) =0.46 P(F and R) =0.32

So then we can create a Venn diagram as we can see on the figure attached.

a. Find the probability that a course has a final exam or a research project.

For this case we can find the probability like this:

P(F UR) = P(F) +P(R) -P(F and R) = 0.72+0.46-0.32=0.86

b. Find the probability that a course has NEITHER of these two requirements.

For this case we can use the complement rule and we can find the probability like this:

P(FUR)' = 1-P(FUR)= 1-0.86 = 0.14

And that's the same value obtained with the diagram.

8 0
3 years ago
Evaluate the following expression 24/3+7•2-15/5+6•2
Diano4ka-milaya [45]

Answer:

31

Step-by-step explanation:

(24/3)+(7x2)-(15/5)+(6*2)

8+(7x2)-(15/5)+(6*2)

8+14-(15/5)+(6*2)

8+14-3+(6*2)

8+14-3+12

22-3+12

19+12

31


Sorry if I did my math wrong :)

6 0
3 years ago
Given: ∠1 ≅ ∠9
Vadim26 [7]

A \: is \: the \: right \: answer

3 0
4 years ago
Read 2 more answers
What is 27 minus 3.204 minus 10.8
lawyer [7]
   27
- 3.204
  10.8
----------

12.996 ?

6 0
3 years ago
Read 2 more answers
A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

velocity = frequency × wavelength   .....................2

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × \rho × g = mg

and

T = v × \rho - v × \rho × g    

from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
3 years ago
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