Answer:
=(k−1)*P(X>k−1) or (k−1)365k(365k−1)(k−1)!
Step-by-step explanation:
First of all, we need to find PMF
Let X = k represent the case in which there is no birthday match within (k-1) people
However, there is a birthday match when kth person arrives
Hence, there is 365^k possibilities in birthday arrangements
Supposing (k-1) dates are placed on specific days in a year
Pick one of k-1 of them & make it the date of the kth person that arrives, then:
The CDF is P(X≤k)=(1−(365k)k)/!365k, so the can obtain the PMF by
P(X=k) =P (X≤k) − P(X≤k−1)=(1−(365k)k!/365^k)−(1−(365k−1)(k−1)!/365^(k−1))=
(k−1)/365^k * (365k−1) * (k−1)!
=(k−1)*(1−P(X≤k−1))
=(k−1)*P(X>k−1)
The problem is modelled in the diagram below
Question 1:
The largest area of the pool is half of the area of the circle
Area of circle = πr², where r is the radius given by 1/2 of the diameter
Area of circle = π(60)² = 11309.73355 feet
Area of half of the circle is 11309.73355/2 = 5654.866... ≈ 5654.87 feet (2dp)
Question b:
The area of the pool is the area of the circle subtracts the area of the triangle.
The area of the circle is 11309.73²
The area of the triangle is 1/2 (60×103.92) = 3117.6 feet²
The area of the pool is 11309.73 - 3117.6 = 7922.13 feet²
The volume of the pool = 7922.13 × 4 = 31688.52 feet³
Note: there isn't any information on the fish tank part so the answer above is assumed for the whole pool.
Answer:
x=2c/3+1
Step-by-step explanation:
Subtract 2c from both sides.
−3x=−3−2c
The equation is in standard form.
−3x=−2c−3
Divide both sides by −3.
-3x/-3=-2c-3/-3
Dividing by −3 undoes the multiplication by −3.
x=-2c-3/-3
Divide −3−2c by −3.
x=2c/3+1
Lets take Gregs weight as “x”. This means that Justins weight is x-15, and x/2 = (x-15)-75.
If we take that last equation, lets combine like terms:
x/2 = x - 15 - 75
x/2 = x - 90
Now multiply both sides by 2 to get rid of the fraction
x = 2x - 180
Subtract 2x from both sides
x - 2x = -180
-x = -180
x = 180 — this is Gregs weight
Justins weight is x-15, so 180-15, which is 165 pounds. Hope this helped.