Answer:
165°
Step-by-step explanation:
The volume of the car is 1800cm^3 which is equivalent to 0.0018m^3
<h3>Scale modelling</h3>
Given the scale factor that model of a car is amde to a scale of 1:40, Giving that the model as 45cm^3, this means that;
1 = 45cm^2
Determine the volume of the car
40 = x
Find the ratio
1/40 = 45/x
x = 40*45
x = 1800cm^3
x = 0.0018m^3
Hence the volume of the car is 1800cm^3 which is equivalent to 0.0018m^3
Learn more on scale factor here; brainly.com/question/25722260
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Let us say that x is the real height of the original hut,
therefore we can establish the equation:
5 : x = 1.5 : 8
5 / x = 1.5 / 8
Solving for x:
x = 5 * 8 / 1.5
x = 40 / 1.5
<span>x = 26.67 ft</span>
The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.
<h3>What is an area bounded by the curve?</h3>
When the two curves intersect then they bound the region is known as the area bounded by the curve.
The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be
Then the intersection point will be given as
Then by the integration, we have
![\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\](https://tex.z-dn.net/?f=%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%5B%20%285%20%5Csin%20%5Ctheta%29%5E2%20-%204%5E2%5D%20d%5Ctheta%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%20%5B25%5Csin%20%5E2%20%5Ctheta%20-%2016%5D%20d%5Ctheta%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cint%20_%7B0.927%7D%5E%7B2.214%7D%20%5B%20%5Cdfrac%7B25%7D%7B2%7D%281%20-%20%5Ccos%202%5Ctheta%20%29%20-%2016%5D%20d%5Ctheta%20%5C%5C)
![\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\](https://tex.z-dn.net/?f=%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5B%5Cdfrac%7B25%20%5Ctheta%20%7D%7B5%7D%20-%20%5Cdfrac%7B25%20%5Ccos%202%5Ctheta%20%7D%7B2%7D%20-%2016%5Ctheta%5D_%7B0.927%7D%5E%7B2.214%7D%20%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Cdfrac%7B1%7D%7B2%7D%20%5B%5Cdfrac%7B25%282.214%20-%200.927%29%20%7D%7B5%7D%20-%20%5Cdfrac%7B25%20%28%5Ccos%202%5Ctimes%202.214%20-%20%5Ccos%202%5Ctimes%200.927%29%20%7D%7B2%7D%20-%2016%282.214%20-%200.927%5D%5C%5C)
On solving, we have
Thus, the area of the region is 3.75 square units.
More about the area bounded by the curve link is given below.
brainly.com/question/24563834
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Answer:
90% CI expects to capture u 90% of time
(a) This means 0.9 * 1000 = 900 intervals will capture u
(b) Here we treat CI as binomial random variable, having probability 0.9 for success
n = 1000
p = 0.9
For this case, applying normal approximation to binomial, we get:
mean = n*p= 900
variance = n*p*(1-p) = 90
std dev = 9.4868
We want to Find : P(890 <= X <= 910) = P( 889.5 < X < 910.5) (integer continuity correction)
We convert to standard normal form, Z ~ N(0,1) by z1 = (x1 - u )/s
so z1 = (889.5 - 900 )/9.4868 = -1.11
so z2 = (910.5 - 900 )/9.4868 = 1.11
P( 889.5 < X < 910.5) = P(z1 < Z < z2) = P( Z < 1.11) - P(Z < -1.11)
= 0.8665 - 0.1335
= 0.733
