Respuesta:
5.101 × 10⁸ km²
Explicación:
Primer paso: Información provista
Radio del planeta Tierra (r): 6371 km
Paso 2: Calcular el área del planeta Tierra
Si suponemos que el planeta Tierra tiene forma esférica, podemos calcular su área (A) usando la siguiente fórmula.
A = 4 × π × r²
A = 4 × π × (6371 km)² = 5.101 × 10⁸ km²
We are given with the following given
dV/dt = 35 m3/min
h = (3/2)D
What is asked is
dr/dt at t = 7 min
The volume of a cone is
V = (1/3)πr²h
From the given
h = (3/2)D
since D = 2r
h = 3r
Subsituting
V = πr³
At t = 7 min, the volume is
V = 35(7)
V = 245
and the radius is
245 = πr³
r = 4.27
Differentiating the volume
dV/dt = 3πr² dr/dt
From the given
dV/dt = 35
Subsituting
35 = 3πr² dr/dt
So,
dr/dt = 0.20 m/s
and
dh/dt = 0.60 m/s
Answer:
15.866%
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 3,282 gallons of kerosene
μ is the population mean = 3,087 gallons of kerosene
σ is the population standard deviation = 195 gallons
For x > 3282 gallons
z = 3282 - 3087/195
z = 1
Probability value from Z-Table:
P(x<3282) = 0.84134
P(x>3282) = 1 - P(x<3282) = 0.15866
Converting to percentage
0.15866 × 100 = 15.866%
The percent of flights on these particular routes that will burn more than 3,282 gallons of kerosene is 15.866%
Answer and Explanation:
Using trig ratios, we can express the given values of sin u and tan v as shown below
![\begin{gathered} \sin u=\frac{opposite\text{ of angle u}}{\text{hypotenuse}}=\frac{2}{5} \\ \tan v=\frac{opposite\text{ of angle v}}{\text{hypotenuse}}=\sqrt[]{21} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csin%20u%3D%5Cfrac%7Bopposite%5Ctext%7B%20of%20angle%20u%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cfrac%7B2%7D%7B5%7D%20%5C%5C%20%5Ctan%20v%3D%5Cfrac%7Bopposite%5Ctext%7B%20of%20angle%20v%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Csqrt%5B%5D%7B21%7D%20%5Cend%7Bgathered%7D)
So we can go ahead and label the sides of the triangle as shown below;
We can find the value of u as shown below;
We can find v as shown below;
