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elena55 [62]
3 years ago
7

How many ways can a person toss a coin 15 times so that the number of heads is between 8 and 12 inclusive

Mathematics
1 answer:
ira [324]3 years ago
6 0

Answer:

16263 ways

Step-by-step explanation:

Given that:

Number of coin tosses, n = 15

Number of heads is between 8 and 12 inclusive

For 8 heads, tails = 15 - 8 = 7

15! / (8!7!) = 6435

For 9 heads, tails = 15 - 9 = 6

15! / (9!6!) = 5005

For 10 heads, tails = 15 - 10 = 5

15! / (10!5!) = 3003

For 11 heads, tails = 15 - 11 = 4

15! / (11!4!) = 1365

For 12 heads, tails = 15 - 8 = 3

15! / (12!3!) = 455

(6435 + 5005 + 3003 + 1365 + 455) = 16,263 ways

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6 0
3 years ago
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gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783, which equation could be used to fi
stepladder [879]
(2x+1)=Jeremy´s age
(2x+3)=Sam´s age.
we suggest this equation:
(2x+1)(2x+3)=783
4x²+6x+2x+3=783
4x²+8x-780=0
x²+2x-195=0

We solve this quadratic equation:
x=[-2⁺₋√(4-4*1*(-195))]/2=(-2⁺₋28)/2

x₁=(-2-28)/2=-15 this solution is not valid.
x₂=(-2+28)/2=13

Jeremy´s age=2x+1=2*13+1=27
Sam´s age=2x+3=2*13+3=29



7 0
3 years ago
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