Answer:
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that the selection of the random pages will contain at least two errors is 0.2644
Step-by-step explanation:
From the information given:
Let q represent the no of typographical errors.
Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let be the random variable that follows a Poisson distribution, then mean
and the mean that the random selection of 50 pages will contain no error is
∴
Pr(q =0) = 0.368
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that 50 randomly page contains at least 2 errors is computed as follows:
P(X ≥ 2) = 1 - P( X < 2)
P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2
P(X ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644
Answer:
Step-by-step explanation:
The three heaviest eggs are between and
Hope this helps and please feel free to comment, ask questions, give feedback, and correct me if I am wrong!
Have a great day!
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Answer:
this applies only if x = <u>-4</u>
Step-by-step explanation:
2x + 1 = 20 + 8x + 5
2x + 1 = 8x + 25 (combine like terms and simplify)
1 = 6x + 25 (subtract 2x from both sides)
-24 = 6x (subtract 25 from both sides)
x = <u>-4</u> (divide 6 from both sides)
Answer:
$3,636.70
Step-by-step explanation:
Given the following:
Initial price of Gold (A) = $2500
Rate of appreciation (r) = 5.5% = 0.055
Worth of gold in 7 years will be?
Period (p) = 7
Using the compound interest formula :
Let F = final amount
F = A( 1 + r/n)^nt
n = number of times Appreciation occurs per period.
Since rate compounds yearly, then, n = 1
F = $2500( 1 + 0.055/1)^(7*1)
F = $2500(1 + 0.055)^7
F = $2500(1.055)^7
F = $2500(1.454679161133794609375)
F = $3636.6979
Amount in 7 years = $3,636.70