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.
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Answer:
32.1 pounds
Step-by-step explanation:
Given data
Percentage of body fat= 19.8%
Weight= 162pounds
Let us find 19.8% of the persons weight
=19.8/100*162
=0.198*162
=32.076
Therefore, 32.1 pounds of the weight is fat
Yes, the sampling distribution is normally distributed because the population is normally distributed.
A sampling distribution is a chance distribution of a statistic obtained from a larger variety of samples drawn from a specific populace. The sampling distribution of a given population is the distribution of frequencies of a variety of various outcomes that would probable occur for a statistic of a populace.
A sampling distribution is a probability distribution of a statistic this is obtained via drawing a huge variety of samples from a particular populace. Researchers use sampling distributions so that you can simplify the technique of statistical inference.
Solution :
mean = μ40
standard deviation σ σ= 3
n = 10
μx = 40
σ x = σ√n = 3/√10 = 0.9487
μ x = 4σ\x = 0.9487
σx = 0.9487
Yes, the sampling distribution is normally distributed because the population is normally distributed.
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Answer:
its 5.795
Step-by-step explanation:
