(Mark 1) Frequency x+2

the population of a colony grows from 250 in the year 2000 to 400 in the year 2004. what is the average rate of change of popula

Ilia_Sergeevich [38]

Given:

Population of a colony in 2000 = 250

Population of a colony in 2004 = 400

To find:

The average rate of change.

Solution:

Let y be the population of the colony at x year.

The population of a colony grows from 250 in the year 2000 to 400 in the year 2004. So, the two points are (2000,250) and (2004,400).

Formula for slope or average rate of change is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{400-250}{2004-2000}$

$m=\dfrac{150}{4}$

$m=37.5$

Therefore, the average rate of change of population is 37.5.

3 0

3 years ago

A 12 foot ladder is leaning against a wall and the foot of the ladder is 5 feet away from the wall. What is the angle between th

Pepsi [2]

Answer:

65.38degrees

Step-by-step explanation:

Given the following

Length of ladder = 12foot = hypotenuse

Distance of the ladder from the wall = 5ft = adjacent

Required

Angle between the ground and the ladder theta

Using the SOH CAH TOA identity

Cos theta = adj/hyp

Cos theta = 5/12

Cos theta = 0.41667

theta = arccos(0.41667)

theta = 65.38 degrees

Hence the angle between the ground and the floor is 65.38degrees

8 0

3 years ago

15 points! The data set shows the weights of pumpkins, in pounds, that are chosen for a photograph in a farming magazine. The re

vladimir2022 [97]

Answer:

Step-by-step explanation:

a). Mean of the given data set = $\frac{32+28+18+40+22}{5}$

= 28

b). x $x-\overline{x}$ $(x-\overline{x})^2$

32 32-28 = 4 16

28 28 - 28 = 0 0

18 18 - 28 = -10 100

40 40 - 28 = 12 144

22 22 - 28 = -6 36

$\sum({x-\overline x})^2$ = 296

c). Standard deviation = $\sqrt{\frac{\sum (x-\overline{x})^2}{n}}$

= $\sqrt{\frac{296}{5} }$

= 7.69

d). Mean of the new set of weights = $\frac{32+33+34+35+36}{5}$

= 34

Since mean of the data set is more close to each data, standard deviation will be less than the deviation given in part (c).

7 0

3 years ago

An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farm

yulyashka [42]

Answer:

$\frac{ds}{dt} = 39.586 km/h$

Step-by-step explanation:

let distance between farmhouse and road is 2 km

From diagram given

p is the distance between road and past the intersection of highway

By using Pythagoras theorem

$s^2 = 2^2 +p^2$

differentiate wrt t

we get

$\frac{d}{dt} s^2 = \frac{d}{dt} (4 + p^2)$

$2s \frac{ds}{dt} =2p \frac{dp}{dt}$

$\frac{ds}{dt} = \frac{p}{s}\frac{dp}{dt}$

$\frac{ds}{dt} = \frac{p}{\sqrt{p^2 +4}} \frac{dp}{dt}$

putting p = 3.7 km

$\frac{ds}{dt} = \frac{3.7}{\sqrt{3.7^2 +4}} 45$

$\frac{ds}{dt} = 39.586 km/h$

3 0

3 years ago

Please explain how you got the answer also, thank you

harina [27]

Answer:

See Explanation

Step-by-step explanation:

You are doing simple math within the square root. For instance where x=-2 you substitute -2 for x then add 2 which gives you 0 leaving you with -6.

for x=2 you end up with "square root 2+2" which gives you 4 and the square root of 4 is 2 so you add -6 to 2 giving you the answer of -4.

7 0

3 years ago