Answer:
Option C: E(X) = 1.65
Step-by-step explanation:
Expected value of X, E(X) = ∑XP(X)
∑ = Summation
X = number of hours
P(X) = probability of a certain number of hours
E(X) = (0*P(X=0)) + (1*P(X=1)) + (2*P(X=2)) + (3*P(X=3)) + (4*P(X=4))
E(X) = (0*0.2) + (1*0.2) + (2*0.4) + (3*0.15) + (4*0.05)
E(X) = 0 + 0.2 + 0.8 + 0.45 + 0.20 = 1.65
E(X) = 1.65 - Option C
Answer:
Step-by-step explanation:
A = P[1 + (r/n)]^(nt)
Look the formula up online to find out what each variable represents if needed.
We are solving for P, which is the principal amount 5 years back.
Substitute 11,000 for A (total amount), 4.5% for r(rate), 2 for n(number of times -- in this case, semiannually), and 5 for t(time -- 5 years).
11000 = P[1+(0.045/2)]^(2x5)
11000 = P (1.0225)^10
11000 = P (1.2492) approx.
Divide both sides by 1.2492 to find P.
P = $8805.64 (approx)
I know that the answer I got is not an option listed. However, I feel that my workings may help other people to answer this question or give them inspiration to answer similar questions. Therefore, I will leave this answer up.
Thanks, and sorry I couldn't do much.
Answer:
9 (approximately)
Step-by-step explanation:
Approximate difference
= pH of basic solution - pH of acidic solution
= 11.2 - 2.4
= 8.8
= 9 (approximately)
Answer: 41.86 miles/hour
Step-by-step explanation:
Use Distance = Speed*Time to find the two missing values with the data that is provided for those two cases [<em>in brackets</em>].
MPH MILES HOURS
50 20 [<em>0.4]</em>
[<em>40</em>] <u>70</u> <u>1.75</u>
Totals 90 miles 2.15 hours
Average = 90 miles/2.15 hours
41.86 mph