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Jet001 [13]
3 years ago
6

Nectar foraging by bumblebees Suppose that, instead of the specific nectar function in Example 2, we have an arbitrary function

N with Ns0d − 0, Nstd > 0, N9std . 0, N0std , 0, and arbitrary travel time T.
(a) Interpret the conditions on the function N.
(b) Show that the optimal foraging time t satisfies the equation N9std − Nstd t 1 T
(c) Show that, for any foraging time t satisfying the equation in part (b), the second derivative condition for a maximum value
Mathematics
1 answer:
Anton [14]3 years ago
7 0

Answer:

B

Step-by-step explanation:

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Omg, I need help! A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6
klemol [59]

Answer:

Average size of the lots = ⅓ acre

Step-by-step explanation:

The question incomplete without specifying what we are to determine.

Question:A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X's. Organize the information in a line plot. What is the average size of the lots? _________ acre

Help anyone?

Solution:

We are asked to organize the information in a line plot. See attachment for the line plot.

Given: 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X'sIn no particular order, the sizes of the lots are:1/6, 1/6, 1/6, 1/6, 1/6, 1/6, 1/3, 1/3, 1/3, 1/2, 1/2, 1/2, 1/2, 1/2 and 1/2 acre.

Let's count the number of lot for each size given.

For 1/6: there are 6 X's on the line plot of 1/6 number of lot for 1/6

= the lot × number of times it occurs

= (1/6) × 6 = 6/6 = 1 acre

For 1/3: there are 3 X's on the line plot of 1/3 number of lot for 1/3 = the lot × number of times it occurs

= (1/3) × 3 = 3/3 = 1 acre

For 1/2: there are 6 X's on the line plot of 1/2 number of lot for 1/2

= the lot × number of times it occurs

= (1/2) × 6 = 6/2= 3 acres

To find average size of the lots, we would sum all lot for each given size then divide by the total number of lots given.Sum of all lot for each given size = 1+1+3

Sum of all lot for each given size = 5

The total number of lots given = 15

Average size of the lots = 5/15 = 1/3

Average size of the lots = ⅓ acre

7 0
3 years ago
PLEASE HELP ONE QUESTION!!!!!! PLEASEEEEE
In-s [12.5K]
VW=CD since they are congruent figure, so CD=6
7 0
3 years ago
Read 2 more answers
Starting at 1:00 pm, the temperature changes -4 degrees Fahrenheit per hour. Write and solve an equation to find how many hours
vesna_86 [32]

Answer:

im guessing it will take 3 hours to get to -1 F

Step-by-step explanation:

4 0
3 years ago
Both Andrew and Karleigh recorded the distance they ran in x minutes on treadmills. Andrew A 2-column table with 2 rows. Column
mars1129 [50]

Answer:

ex:Speed (S) is the ratio of the distance (D) covered to the time (t) taken.

That is, S = D/t

Suppose Andrew ran a distance D1 in 1 hour (3600 seconds) at a Speed, say S1, we have

S1 = D1/t

We can then say he ran a distance

D1 = t × S1

= 3600S1

Similarly, let's say Karleigh ran a distance

D2 = t × S2

= 3600S2

Let us compare these two, you will notice that the bigger number between S1 and S2 is going to determine the bigger number between D1 and D2.

Let's choose random numbers for S1 and S2 for clarity, say S1 = 5, S2 = 10

D1 = 3600 × 5

= 18000

D2 = 3600 × 10

= 36000

This makes D2 bigger than D1. this is an example i found on the internet.

Step-by-

hope this helps, good luck

7 0
3 years ago
Read 2 more answers
Use the model below to estimate the average annual growth rate of a certain country's population for 1950, 1988, and 2010, where
Morgarella [4.7K]

Answer:

The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.

Step-by-step explanation:

The given equation is

Y=-0.0000084x^3+0.00211x^2-0.205x+8.423

Where Y is the annual growth rate of  a certain country's population and x is the number of years after 1900.

Difference between 1950 and 1900 is 50.

Put x=50 in the given equation.

Y=-0.0000084(50)^3+0.00211(50)^2-0.205(50)+8.423

Y=2.398

Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.

Difference between 1988 and 1900 is 88.

Put x=88 in the given equation.

Y=-0.0000084(88)^3+0.00211(88)^2-0.205(88)+8.423

Y=0.9984752\approx 0.9985

Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.

Difference between 2010 and 1900 is 110.

Put x=110 in the given equation.

Y=-0.0000084(110)^3+0.00211(110)^2-0.205(110)+8.423

Y=0.2236

Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.

8 0
3 years ago
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