First, we have
s1/r1 = s2/r2
The question also states the fact that
s/2πr = θ/360°
Rearranging the second equation, we have
s/r = 2πθ/360°
Then we substitute it to the first equation
s1/r1 = 2πθ1/360°
s2/r2 = 2πθ2/360°
which is now
2πθ1/360° = 2πθ2/360°
By equating both sides, 2π and 360° will be cancelled, therefore leaving
θ1 = θ2
The problem is asking how much each person will need to pay. Simplifying the problem into an equation with variables (an algorithm) will greatly help you solve it:
S = Sales Tax = $ 7.18 per any purchase
A = Admission Ticket = $ 22.50 entry price for one person (no tax applied)
F = Food = $ 35.50 purchases for two people
We know the cost for one person was: (22.50) + [(35.50/2) + 7.18] =
$ 47.43 per person. Now we can check each method and see which one is the correct algorithm:
Method A)
[2A + (F + 2S)] / 2 = [ (2)(22.50) + [35.50 + (2)(7.18)] ]/ 2 = $47.43
Method A is the correct answer
Method B)
[(2A + (1/2)F + 2S) /2 = [(2)(22.50) + 35.50(1/2) + (2)7.18] / 2 = $38.55
Wrong answer. This method is incorrect because the tax for both tickets bought are not being used in the equation.
Method C)
[(A + F) / 2 ]+ S = [(22.50 + 35.50) / 2 ] + 7.18 = $35.93
Wrong answer. Incorrect Method. The food cost is being reduced to the cost of one person but admission price is set for two people.
Distinct prime factors are factors that can not be reduced any further.... Such factors are like 3,5, and 7.
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
There won’t be enough for each friend
