The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the

Answer:

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Step-by-step explanation:

a) To find P(X=2), we calculate: P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4) which equals 1 - 0.41 - 0.18 - 0.06 - 0.06, resulting in 0.29

b) For P(X<2), we sum P(X=0) and P(X=1): 0.41 + 0.18 yields 0.59

c) To obtain P(X≤2), we add P(X=0), P(X=1), and P(X=2): 0.41 + 0.18 + 0.29 equals 0.88

d) To calculate P(X>2), we find P(X=3) + P(X=4): 0.06 + 0.06 gives us 0.12

e) For P(X=1 or X=4), we use the union of probabilities: P(X=1) + P(X=4) which is 0.18 + 0.06, resulting in 0.24

f) P(1≤X≤4) is found by adding P(X=1), P(X=2), P(X=3), and P(X=4): 0.18 + 0.29 + 0.06 + 0.06 results in 0.59

Answer:

600 books

Step-by-step explanation:

The dimensions of the bin are

5 by 2 by 3

The volume of the bin is found by multiplying these three dimensions.

Volume of Bin = 5 * 2 * 3 = 30 cubic feet

To find the volume of each book, we use the same method. The dimensions of one book are:

1 by 0.5 by 0.1

Volume of 1 book = 1 * 0.5 * 0.1 = 0.05 cubic feet

The total number of books fitting into the bin is calculated by:

30/0.05 = 600 books

Response:

24.2 (rounded to the nearest tenth)

Detailed breakdown:

The data type involved is ungrouped

standard deviation =√ [∑ (x-μ)² / n]

average (μ)=∑x/n

           =

          =33.3

For the values 5, 5, 6, 12, 13, 26, 37, 49, 51, 56, 56, 84, the x-μ results will be

                   -28.3, -28.3, -27.3, -21.3, -7.3, 3.7, 15.7,17.7, 22.7,22.7,50.7

Calculating (x-μ)² gives: 800.89, 800.89, 745.29, 453.69, 53.29, 13.69, 246.49, 313.29, 515.29, 515.29, 2570.49

Then, ∑ (x-μ)² results in = 7028.59

Thus, standard deviation = √(7028.59 / 12)

                      =24.2

Answer:

link:

Step-by-step explanation: