Shown again is the table indicating the marital status of the U.S. population in 2015. Numbers in the table are expressed in millions. Use the data in the table to solve Exercises 65-76. Express probabilities as simplified fractions and as decimals rounded to the nearest hundredth. MARITAL STATUS OF THE U.S. POPULATION, AGES 15 OR OLDER, 2015, IN MILLIONS Married Never Married Divorced Widowed Total Male 66 43 11 3 123 Female 67 38 15 11 131 Total 133 81 26 14 254 If one person is selected from the population described in the table, find the Probability, that the person 68. has never been married or is divorced. 107 254 ≈ 0.42
Shown again is the table indicating the marital status of the U.S. population in 2015. Numbers in the table are expressed in millions. Use the data in the table to solve Exercises 65-76. Express probabilities as simplified fractions and as decimals rounded to the nearest hundredth. MARITAL STATUS OF THE U.S. POPULATION, AGES 15 OR OLDER, 2015, IN MILLIONS Married Never Married Divorced Widowed Total Male 66 43 11 3 123 Female 67 38 15 11 131 Total 133 81 26 14 254 If one person is selected from the population described in the table, find the Probability, that the person 68. has never been married or is divorced. 107 254 ≈ 0.42
Solution Summary: The author explains the formula used to calculate the probability that the person has never been married or is divorced.
Shown again is the table indicating the marital status of the U.S. population in 2015. Numbers in the table are expressed in millions. Use the data in the table to solve Exercises 65-76. Express probabilities as simplified fractions and as decimals rounded to the nearest hundredth.
MARITAL STATUS OF THE U.S. POPULATION, AGES 15 OR OLDER, 2015, IN MILLIONS
Married
Never Married
Divorced
Widowed
Total
Male
66
43
11
3
123
Female
67
38
15
11
131
Total
133
81
26
14
254
If one person is selected from the population described in the table, find the Probability, that the person
68. has never been married or is divorced.
107
254
≈
0.42
=
Q6 What will be the allowable bearing capacity of sand having p = 37° and ydry
19 kN/m³ for (i) 1.5 m strip foundation (ii) 1.5 m x 1.5 m square footing and
(iii)1.5m x 2m rectangular footing. The footings are placed at a depth of 1.5 m
below ground level. Assume F, = 2.5. Use Terzaghi's equations.
0
Ne
Na
Ny
35 57.8 41.4 42.4
40 95.7 81.3 100.4
Q1 The SPT records versus depth are given in table below. Find qan for the raft 12%
foundation with BxB-10x10m and depth of raft D-2m, the allowable
settlement is 50mm.
Elevation, m 0.5 2
2 6.5 9.5 13 18 25
No.of blows, N 11 15 29 32 30 44
0
estigate shear
12%
2
Chapter 11 Solutions
Thinking Mathematically Plus MyLab Math -- Access Card Package (7th Edition) (What's New in Service Math)
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