so the first term is 36500.
after a year, he gets 2375 extra, so
1st year, 36500
2nd year 36500 + 2375
3rd year 36500 + 2375 + 2375
4th year 36500 + 2375 + 2375 + 2375
and so on
so the common difference is 2375, namely the number we add in order to get the following term.
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=36500\\ d=2375 \end{cases} \\\\\\ a_n=36500+(n-1)2375\implies a_n=36500+2375(n-1) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{his salary after 15 years, \boxed{n = 15}}}{a_{15}=36500+2375(15-1)}\implies a_{15}=36500+2375(14) \\\\\\ a_{15}=36500+33250\implies a_{15}=69750](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D36500%5C%5C%20d%3D2375%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D36500%2B%28n-1%292375%5Cimplies%20a_n%3D36500%2B2375%28n-1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhis%20salary%20after%2015%20years%2C%20%5Cboxed%7Bn%20%3D%2015%7D%7D%7D%7Ba_%7B15%7D%3D36500%2B2375%2815-1%29%7D%5Cimplies%20a_%7B15%7D%3D36500%2B2375%2814%29%20%5C%5C%5C%5C%5C%5C%20a_%7B15%7D%3D36500%2B33250%5Cimplies%20a_%7B15%7D%3D69750)
Answer:
7/25
Step-by-step explanation:
Probability is the likelihood or chance that an event will occur.
Probability = expected outcome/total outcome
If the box contains 3 black pens, 7 blue pens, and 5 red pens, the total number of pens in the box is 3+7+5 = 15pens
Probability of pulling a red pen and replacing = 5/15 = 1/3
Probability of pulling a blue pen and replacing = 7/15
Probability of pulling a black pen and not replacing(using it) = 3/15
{Note that the total number of outcome used will still be 15 since the black pen was pulled out last even though it wasn't replaced.}
P(red, then blue, then black) = 1/3×7/15×3/15
= 63/225
= 7/25
Answer. :
True
Explanation :
The fraction 4/5 is equivalent to 24/30 (To get this I multiplied both the numerator and denominator by 6). The fraction 5/6 is equivalent to 25/30 (To get this I multiplied both the numerator and denominator by 5). ... Since 25 is greater than 24, 25/30 (or originally 5/6) is the larger fraction.
The correct answer would be C
Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
