Answer:
0.9953, 3.3629<x<4.0371
Step-by-step explanation:
Given that slader the internal revenue service claims it takes an average of 3.7 hours to complete a 1040 tax form, assuming th4e time to complete the form is normally distributed witha standard devait of the 30 minutes:
If X represents the time to complete then
X is N(3.7, 0.5) (we convert into uniform units in hours)
a) percent of people would you expect to complete the form in less than 5 hours
=
b) P(b<x<c) = 0.50
we find that here
c = 4.0371 and
b = 3.3629
Interval would be
The length of an arc is the fraction of its circumference based on the given intercepted angle. This is given by the equation,
L (arc) = (Angle / 360) x 2πr
Substituting the known values,
L (arc) = (40 / 360) x 2π(8 inch) = 16π/9 inch
Thus, the length of the arc is approximately equal to 5.585 inches.
Answer:
-1
Step-by-step explanation:
-7/10+7/15+1/-20+-9/10+11/15+11/-20
= -7/10 + -9/10 + 7/15 + 11/15 + 1/-20 + 11/-20
= -16/10 + 18/15 + 12/-20
= -8/5 + 6/5 + -3/5
= (-8+6-3)/5
= -5/5
= -1
It's 24k-6
By order of operations, you do the multiplication first: 3k(4) = 12k
12k-6+12k now combine like terms
24k-6
Hope that helps
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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