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2 years ago

9

A researcher who was conducting a study caught, tagged, and released 95 opossums. Weeks later, the researcher caught a total of

170 opossums, of which 19 were tagged. To the nearest whole number, what is the best estimate for the opossum population?

1 answer:

Mamont248 [21]2 years ago

6 0

Answer:

Your answer would be 850

Step-by-step explanation:

You might be interested in

What are the two ratios that are equivalent to 2:9

weeeeeb [17]

Answer:

bc

Step-by-step explanation:

2:9 is equivalent to

18: 81 because is 2*9 : 9*9

12:54 because is 2*6 : 9* 6

8 0

2 years ago

You plant a garden in the shape of a triangle as shown in the figure. What is the perimeter of the garden? Find the values of x

Vadim26 [7]

Answer:

x = 42°

y = 69°

Step-by-step explanation:

From the picture attached,

Given triangle is an isosceles triangle.

Therefore, two sides will be equal and measure 8 yd.

Measure of 3rd side of the triangle = 6 yd

Perimeter of the garden = 8 + 8 + 6

= 22 yd

In an isosceles triangle, opposite angles of the equal sides will be equal.

Therefore, y = 69°

By the property of interior angles of a triangle,

x° + y° + 69° = 180°

x° + 69° + 69° = 180°

x = 180 - 138

x = 42°

7 0

2 years ago

Newton's law of cooling is:

Mnenie [13.5K]

Answer:

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

We can use the substitution $w = u-T$ and $dw =du$ so then we have:

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

$ln |w| = -kt +C$

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

And if we replace w = u-T we got:

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

For this case we know that $k =-0.15 hr$ since w ehave a cooloing, $T_{i}= 70 F, T_{amb}=11F$ , we have this model:

$u(t) = (70-11) e^{-0.15t} +11$

And if we want that the temperature would be 32F we can solve for t like this:

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

7 0

3 years ago

What is 128 5/8×2?????? plz helppp

olya-2409 [2.1K]

Answer:

257.25

Hope This Helps! Have A Nice Day!!

8 0

2 years ago

Read 2 more answers

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra

bearhunter [10]

Answer:

$\frac{dh}{dt}=-\frac{1}{2\pi}cm/min$

Step-by-step explanation:

From similar triangles, see diagram in attachment

$\frac{r}{4}=\frac{h}{16}$

We solve for $r$ to obtain,

$r=\frac{h}{4}$

The formula for calculating the volume of a cone is

$V=\frac{1}{3}\pi r^2h$

We substitute the value of $r=\frac{h}{4}$ to obtain,

$V=\frac{1}{3}\pi (\frac{h}{4})^2h$

This implies that,

$V=\frac{1}{48}\pi h^3$

We now differentiate both sides with respect to $t$ to get,

$\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}$

We were given that water is drained out of the tank at a rate of $2cm^3/min$

This implies that $\frac{dV}{dt}=-2cm^3/min$ .

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means $h=8cm$ .

We substitute this values to obtain,

$-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}$

$\Rightarrow -2=4\pi \frac{dh}{dt}$

$\Rightarrow -1=2\pi \frac{dh}{dt}$

$\frac{dh}{dt}=-\frac{1}{2\pi}$

3 0

3 years ago

Read 2 more answers