How would the author of Passage 1

A hotel manager recorded the percentage of rooms that were occupied each day over a period of 25 days. The data she collected is

inn [45]

Answer:

15

Explanation:

plato users!

3 0

2 years ago

PROS AND CONS OF ARRIVE ALIVE CAMPAIGN

blsea [12.9K]

<h3>ANSWER:</h3>

Pros of Arrive Alive Campaign are:

Mainly it causes you and your co-individuals to be protected with no brutal mishaps.
Whereas it additionally helps your driving abilities and to keep up your vehicle for more.
Following the standard will assist you with helping focus on temporary routes, construction work, and dangers.
These guidelines will spare your life and furthermore it will keep your family glad and heavenly.
Following the traffic, rules guarantee you as a responsible citizen.
Mainly while following the guidelines overwhelming traffic will be diminished and the travelling time will likewise get decrease.

Cons of Arrive Alive Campaign:

The just con of following the traffic guideline is during the red sign, particularly for emergency vehicles, which stalled out between the substantial traffic.

5 0

4 years ago

a helicopter flies parallel to the ground at an altitude of 1/2 kilometer and at a speed of 2 kilometers per minute. if the heli

Murrr4er [49]

The Pythagoras' theorem and the uniform motion allows to find the answer for the rate of change of distance is

v = 32.3 m / s

The distance is the length of the segment that a white house wing with the helicopter, this can be found using the Pythagoras' theorem

R = $\sqrt{x^2 + y^2}$

Where R is the distance, x is the horizontal distance where the helicopter flies and y is the vertical distance that corresponds to the flight height

y = 500 m

Since the helicopter flies at constant speed, we can use the uniform motion relation

$v_h$ = $\frac{x}{t}$

x = $v_h$ t

Where $v_h$ is the speed of the helicopter and t is the time

We substitute

r = $\sqrt {(V_h \ t)^2 + y^2}$

Kinematics defines velocity as the change of position in the unit of time

$v = \frac{dR}{dt}$

$v = \frac{1}{2} \ \frac{2 ( v_h t) v_h}{ \sqrt{(v_h t)^2 + y^2} }$

$v = \frac{v_h^2 \ t}{ \sqrt{(v_h t)^2 + y^2} }$

This expression is the change in distance as a function of time for a given speed. In the exercise, this speed is requested for a time of one minute, for the helicopter speed of

v = 2km/min ( $\frac{1000m}{1km}$ ) ( $\frac{1 min}{60s}$ ) = 33.33 m / s