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Delvig [45]
3 years ago
12

baa baa the black sheep had 7 pounds of wool if he separated the wool equally into 3 bags how much wool would be in 2bags

Mathematics
1 answer:
Andreyy893 years ago
8 0

Answer:

4 pounds and 2/3 of a pound

each bag would contain 2 pounds and 1/3 of a pound so two pounds would be two times that so 4 pounds and 2/3 of a pound is the answer.

i.dk if this is what u needed but good luck

hope this helped

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Which number is closest to √18
Mars2501 [29]

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4.2 or 4

Step-by-step explanation:

√18 = 4.24264068712

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or 4

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The diagram below shows a square cardboard with an area of 132.25 cm ².A square of side x cm is cut and removed from each of its
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In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 40,000 dollars. Othe
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Answer:

We conclude that  expected value of this game is -0.865$.

Step-by-step explanation:

We know that in a lottery game, a player picks six numbers from 1 to 27.

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C_6^{27}=296010

As there is only one advantageous combination, we conclude that the number of non-winning combinations is 296009.

He can win 40,000 dollars.

We calculate:

E(X)=\frac{1}{296010}\cdot 40000\$- \frac{296009}{296010}\cdot 1\$\\\\E(X)=\frac{40000-296009}{296010}\, \$\\\\E(X)=-0.865\, \$

We conclude that  expected value of this game is -0.865$.

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What is equavalent to one ninth
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Read 2 more answers
Find the area of the shaded region
Alex777 [14]

∆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

\sin(30^\circ) = \dfrac{h}{4\,\rm cm} \implies h= 2\,\rm cm

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

\left(\dfrac b2\right)^2 + h^2 = (4\,\mathrm{cm})^2 \implies b = 4\sqrt3 \,\rm cm

where b 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

7 0
2 years ago
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