we have
Step 
<u>Clear the variable y</u>
Adds 
both sides
Step 
<u>Convert in function notation</u>
Let
therefore
<u>the answer is</u>
For this problem, we plug in the <em>numbers for t</em> and the <em>appropriate letters for e</em> to the equation <em>e = 300 - 10t</em>.
a) a = 300 - 10(-2)
simplify: a = 300 + 20
simplify: a = 320 ft.
But, this is not a viable point because when t≤0, she doesn't move anywhere, thus, should consistently be 300 ft.
b) b = 300 - 10(3.5)
simplify: b = 300 - 35
simplify: b = 265 ft.
c) c = 300 - 10(30)
simplify: c = 300 - 300
simplify: c = 0 ft.
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)
A(bx − c) ≥ bc, implies (bx − c) ≥ bc /a and then bx ≥ bc/a + c, x<span>≥ c/a +c/b
so the solution is </span><span>3. [c/a + c/b, infinity)</span>
