X=17+(sqrt(2y)/2),x=17-(sqrt(2y)/2) i believe is the answer
Answer:
V = 34,13*π cubic units
Step-by-step explanation: See Annex
We find the common points of the two curves, solving the system of equations:
y² = 2*x x = 2*y ⇒ y = x/2
(x/2)² = 2*x
x²/4 = 2*x
x = 2*4 x = 8 and y = 8/2 y = 4
Then point P ( 8 ; 4 )
The other point Q is Q ( 0; 0)
From these two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.
Now with the help of geogebra we have: In the annex segment ABCD is dy then
V = π *∫₀⁴ (R² - r² ) *dy = π *∫₀⁴ (2*y)² - (y²/2)² dy = π * ∫₀⁴ [(4y²) - y⁴/4 ] dy
V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴
V = π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )
V = π * [256/3 - 51,20]
V = 34,13*π cubic units
5 will go into 515, 103 times.
Answer:
a=4220-2t
This would be if we are finding the altitude of the submarine from the surface of the water(how many meters away from the surface it is). I don't know what it would be if you want to know from the bottom of the ocean.
Answer:
Step-by-step explanation:
Hello!
You have the information for two variables
X₁: Number of consumer purchases in France that were made with cash, in a sample of 120.
n₁= 120 consumer purchases
x₁= 48 cash purchases
p'₁= 48/120= 0.4
X₂: Number of consumer purchases in the US that were made with cash, in a sample of 55.
n₂= 55 consumer purchases
x₂= 24 cash purchases
p'₂= 24/55= 0.4364
You need to construct a 90% CI for the difference of proportions p₁-p₂
Using the central limit theorem you can approximate the distribution of both sample proportions p'₁ and p'₂ to normal, so the statistic to use to estimate the difference of proportions is an approximate standard normal:
[(p'₁-p'₂) ±
*
]

[(0.4-0.4364)±1.648 *
]
[-0.1689;0.0961]
The interval has a negative bond, it is ok, keep in mind that even tough proportions take values between 0 and 1, in this case, the confidence interval estimates the difference between the two proportions. It is valid for one of the bonds or the two bonds of the CI for the difference between population proportions to be negative.
I hope this helps!