Answer:
Step-by-step explanation:
The 20th of February for seven nights:
Mr. Jones: 714
Mrs. Jones: 714
Child 1: 95% of 714 = 678.30
Child 2: 95% of 714 = 678.30
Total Price: 2784.60
The 10th of April for fourteen nights:
Mr. Jones: 802
Mrs. Jones: 802
Child 1: 85% of 802 = 681.70
Child 2: 85% of 802 = 681.70
Total Price: 2967.40
The 10th of April for fourteen nights is 182.80 pounds greater than the 20th of February for seven nights.
Answer:
Union President Abraham Lincoln set the Union's first naval goal when he declared a blockade of the Southern coasts. His plan was to cut off Southern trade with the outside world and prevent sale of the Confederacy's major crop, cotton.Union President Abraham Lincoln set the Union's first naval goal when he declared a blockade of the Southern coasts. His plan was to cut off Southern trade with the outside world and prevent sale of the Confederacy's major crop, cotton.Union President Abraham Lincoln set the Union's first naval goal when he declared a blockade of the Southern coasts. His plan was to cut off Southern trade with the outside world and prevent sale of the Confederacy's major crop, cotton.Union President Abraham Lincoln set the Union's first naval goal when he declared a blockade of the Southern coasts. His plan was to cut off Southern trade with the outside world and prevent sale of the Confederacy's major crop, cotton.
Step-by-step explanation:
The tree diagram for the probability is shown below
P(Clay|Positive) is read 'Probability of Clay given the result is Positive'.
This is a case of conditional probability.
The formula for conditional probability is given as
P(Clay|Positive) = P(Clay∩Positive) ÷ P(Positive)
P(Clay∩Positive) = 0.21×0.48 = 0.1008
P(Positive) = P(Rock∩Positive) + P(Clay∩Positive) + P(Sand∩Positive)
P(Positive) = (0.53×0.53) + (0.21×0.48) + (0.26×0.75)
P(Positive) = 0.2809 + 0.1008 + 0.195
P(Positive) = 0.5767
Hence,
P(Clay|Positive) = 0.1008÷0.5767 = 0.175 (rounded to 3 decimal place)
∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find

where
is the length of the altitude originating from vertex O, and so

where
is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²
Multiplicative identity, which states that anything multiplied by 1 is itself.