Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Assume the mower moves wit
1 answer:
Answer:
t = 11.9 h = 0.49 day
Explanation:
First of all, we will find the area covered by the mower per unit time:
where,
v = speed of mower = 1 km/h = 1000 m/h
w = width of mower = 0.6 m
Therefore,
Now, the standard football ground has an area of:
A = 105 m x 68 m = 7140 m²
Therefore, The time required to mow the lawn will be:
<u>t = 11.9 h = 0.49 day</u>
You might be interested in
Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate
The minimum power required (at 100% efficiency) is
The actual power will be higher because 100% efficiency is not possible.
Answer: 339.8 W (nearest tenth)
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate
The minimum power required (at 100% efficiency) is
The actual power will be higher because 100% efficiency is not possible.
Answer: 339.8 W (nearest tenth)
The area under the standard normal curve between z=0.19 and z=2.18 is,
P(0.19<z<2.18)= 0.4101
<h3>How can we calculate the area under the curve?</h3>To calculate the The area under the standard normal curve between z=0.19 and z=2.18, we are using two things,
<u>Step 1</u>: The formula,
P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)
<u>Step 2</u>: The statistical values of the area under the curve we get from the picture.
Now we put the known values from the picture in the above formula, we get,
P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)
Or, P(0.19<z<2.18)= 0.9854-0.5753
Or, P(0.19<z<2.18)= 0.4101
From the above calculation we can easily conclude that,
The area under the standard normal curve between z=0.19 and z=2.18 is,
P(0.19<z<2.18)= 0.4101
Learn more about the standard normal curve:
#SPJ4
