Option A: The subjects are numbered 1–50 and these numbers are put into a random number generator. The first 25 random numbers, ignoring repeats, represent the subjects assigned to the new model group. The remaining 25 subjects will wear the current model.
Step-by-step explanation:
We are told that Twenty - five runners will receive a pair of the new model and the remaining 25 will receive a pair of the current model.
Now, a completely randomized design is defined as one where the treatments will be assigned completely at random in such a manner that each experimental unit will posses the same chance of receiving any of the treatments given.
To this effect, the most likely answer will be picking people at random to assign either the new model or current model.
This is usually done by making a list of the treatment combinations and then assigning a random number to each subject.
Looking at the options, the most appropriate will be option A.
Option A: The subjects are numbered 1–50 and these numbers are put into a random number generator. The first 25 random numbers, ignoring repeats, represent the subjects assigned to the new model group. The remaining 25 subjects will wear the current model.
Step-by-step explanation:
We are told that Twenty - five runners will receive a pair of the new model and the remaining 25 will receive a pair of the current model.
Now, a completely randomized design is defined as one where the treatments will be assigned completely at random in such a manner that each experimental unit will posses the same chance of receiving any of the treatments given.
To this effect, the most likely answer will be picking people at random to assign either the new model or current model.
This is usually done by making a list of the treatment combinations and then assigning a random number to each subject.
Looking at the options, the most appropriate will be option A.
7 of 100 parts have been shaded, and 7 hundredths is written as 0.07
Step-by-step explanation:
Note that we have a grid of 100 squares and 7 of those squares are shaded. We can represent the shaded portion by saying 7/100 which is equivalent to .07.
With this information, we can say 7 out of 100 is shaded and is representative of 7 hundredths.
Note that each digit after the decimal is a multiple of (1/10). We can express this as (1/10)^n where n is the digit place after the decimal.
In this case, we have 7, two digits after the decimal so it will be equivalent to 7 * (1/10)^2 = 7 * (1/100) = 7/100.
Cheers.
1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s
Explanation:
1) First Scenario.
After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.
Before starting, let's reduce the magnitudes to the SI system
v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s
y = 2650 m
Let's start by looking for the time it takes for the load to reach the ground.
y = y₀ + v_{oy} t - ½ g t²
in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t2
t =
t = √(2 2650/9.8)
t = 23.26 s
this is the horizontal scrolling time
x = v₀ t
x = 366.67 23.26
x = 8527 m
the speed at the point of arrival is
v_y = v_{oy} - g t = 0 - gt
v_y = - 9.8 23.26
v_y = -227.95 m / s
Module and angle form
v =
v = √(366.67² + 227.95²)
v = 431.75 m / s
θ = tan⁻¹ (v_y / vₓ)
θ = tan⁻¹ (227.95 / 366.67)
θ = - 31.97º
measured clockwise from x axis
We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.
2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º
we look for the components of speed
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v₀ sin θ
we look for the time for the arrival point that has coordinates x = 0, y = 0
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + vo sin θ t - ½ g t²
0 = -50 + vo sin 50 t - ½ 9.8 t²
x = x₀ + v₀ₓ t
0 = x₀ + vo cos θ t
0 = -400 + vo cos 50 t
podemos ver que tenemos un sistema de dos ecuación con dos incógnitas
50 = 0,766 vo t – 4,9 t²
400 = 0,643 vo t
resolved
50 = 0,766 ( ) t – 4,9 t²
50 = 476,52 t – 4,9 t²
t² – 97,25 t + 10,2 = 0
we solve the quadratic equation
t = [97.25 ± ] / 2
t = 97.25 ±97.04] 2
t₁ = 97.145 s
t₂ = 0.1 s≈0
the correct time is t1 the other time is the time to the launch point,
t = 97.145 s
let's find the initial velocity
x = x₀ + v₀ cos 50 t
0 = -400 + v₀ cos 50 97.145
v₀ = 400 / 62.44
v₀ = 6.4 m / s
The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation. ... The cycling of water in and out of the atmosphere is a significant aspect of the weather patterns on Earth .
The water cycle on Earth
The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.
Together, these five processes - condensation, precipitation, infiltration, runoff, and evapotranspiration- make up the Hydrologic Cycle. Water vapor condenses to form clouds, which result in precipitation when the conditions are suitable.
Explanation:
1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s
Explanation:
1) First Scenario.
After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.
Before starting, let's reduce the magnitudes to the SI system
v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s
y = 2650 m
Let's start by looking for the time it takes for the load to reach the ground.
y = y₀ + v_{oy} t - ½ g t²
in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t2
t =
t = √(2 2650/9.8)
t = 23.26 s
this is the horizontal scrolling time
x = v₀ t
x = 366.67 23.26
x = 8527 m
the speed at the point of arrival is
v_y = v_{oy} - g t = 0 - gt
v_y = - 9.8 23.26
v_y = -227.95 m / s
Module and angle form
v =
v = √(366.67² + 227.95²)
v = 431.75 m / s
θ = tan⁻¹ (v_y / vₓ)
θ = tan⁻¹ (227.95 / 366.67)
θ = - 31.97º
measured clockwise from x axis
We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.
2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º
we look for the components of speed
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v₀ sin θ
we look for the time for the arrival point that has coordinates x = 0, y = 0
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + vo sin θ t - ½ g t²
0 = -50 + vo sin 50 t - ½ 9.8 t²
x = x₀ + v₀ₓ t
0 = x₀ + vo cos θ t
0 = -400 + vo cos 50 t
podemos ver que tenemos un sistema de dos ecuación con dos incógnitas
50 = 0,766 vo t – 4,9 t²
400 = 0,643 vo t
resolved
50 = 0,766 ( ) t – 4,9 t²
50 = 476,52 t – 4,9 t²
t² – 97,25 t + 10,2 = 0
we solve the quadratic equation
t = [97.25 ± ] / 2
t = 97.25 ±97.04] 2
t₁ = 97.145 s
t₂ = 0.1 s≈0
the correct time is t1 the other time is the time to the launch point,
t = 97.145 s
let's find the initial velocity
x = x₀ + v₀ cos 50 t
0 = -400 + v₀ cos 50 97.145
v₀ = 400 / 62.44
v₀ = 6.4 m / s
The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation. ... The cycling of water in and out of the atmosphere is a significant aspect of the weather patterns on Earth .
The water cycle on Earth
The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.
Together, these five processes - condensation, precipitation, infiltration, runoff, and evapotranspiration- make up the Hydrologic Cycle. Water vapor condenses to form clouds, which result in precipitation when the conditions are suitable.
Explanation:
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