First you need to get all variables on one side. You do that by multiplying the reciprocal of the fraction to everything.
N(5/1) x 1/5(5/1)=2/15(5/1)
5n=2/15 x 5/1
Then solve the multiplication problem
5n= 10/15
Then you should reduce the fraction
5n=2/3
Then divide both sides by 5
5n/5=N
2/3 divided by 5 =2/15
N=2/15
Answer:
5.3
Step-by-step explanation:
Cos(68) = adjacent/hypotenuse
Cos(68) = 2/AC
AC = 2/cos(68)
AC = 5.3389343253
AC = 5.3
For this case we use the following formula
Area of Sector = Area * radians of sector / 2 * pi radians
Where,
Area: it is the area of the complete circle.
We have then:
Area = pi * r ^ 2
Area = pi * (6) ^ 2
Area = 36pi
Substituting values:
5pi = 36pi * radians of sector / 2 * pi
Clearing:
radians of sector = ((5pi) * (2pi)) / (36pi)
radians of sector = (10pi ^ 2) / (36pi)
radians of sector = (10pi) / (36)
radians of sector = (10/36) pi
radians of sector = (5/18) pi
in degrees:
(5/18) pi * (180 / pi) = 50 degrees
Answer:
The measure of the central angle is:
50 degrees
The required probability is ![\frac{36}{55}](https://tex.z-dn.net/?f=%5Cfrac%7B36%7D%7B55%7D)
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, ![\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20probability%20%7D%3D%5Cfrac%7B%5Ctext%20%7B%20favourable%20outcomes%20%7D%7D%7B%5Ctext%20%7B%20total%20outcomes%20%7D%7D)
Probability for first draw to be non-defective ![=\frac{11-2}{11}=\frac{9}{11}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B11-2%7D%7B11%7D%3D%5Cfrac%7B9%7D%7B11%7D)
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective ![=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B10-2%7D%7B10%7D%3D%5Cfrac%7B8%7D%7B10%7D%3D%5Cfrac%7B4%7D%7B5%7D)
(printers after first slot = 10; total defective printers = 2)
Then, total probability ![=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B9%7D%7B11%7D%20%5Ctimes%20%5Cfrac%7B4%7D%7B5%7D%3D%5Cfrac%7B36%7D%7B55%7D)