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Debora [2.8K]
3 years ago
5

When we toss a penny, experience shows that the probability (long term proportion) of a head is close to 1-in-2. suppose now tha

t we toss the penny repeatedly until we get a head. what is the probability that the first head comes up in an odd number of tosses (one, three, five, and so on)?
Mathematics
1 answer:
Nana76 [90]3 years ago
7 0
<span>2/3 or 0.66666
       
This is a sum of an infinite series problem. A sequence of 1 will happen with a probability of 0.5 A sequence of 3 will happen with a probability of 1/2^3, 1/8, = 0.125 In general we have an infinite series of 1/2^1 + 1/2^3 + 1/2^5 + ... + 1/2^(2n-1) where n >= 1 The sum of such a series with a constant ratio between sequential terms is S = s1/(1-r) where s1 = first term in the series r = ratio between terms. The value for s1 = 0.5 as shown above and the 2nd term is 0.125. So r = 0.125 / 0.5 = 0.25 And the sum of the infinite series is S = s1/(1-r) S = 0.5/(1 - 0.25) S = 0.5/0.75 S = 2/3 S = 0.666..66 So the probability of the first head coming up in an odd number of tosses is 2/3, or 66.6%</span>
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Peyton received a graduation gift from his grandparents of $12,700. He
arlik [135]

Answer:

Peyton's account will have $13,842.18 after a year.

Step-by-step explanation:

Given that Peyton received $ 12,700 and decided to invest it for a year in an account that grants an interest of 8.8% per year, compounded semiannually, to determine the amount of money that will be in said account after the passage of one year, it is necessary to perform the following calculation:

X = 12,700 (1 + 0.088 / 2) ^ 1x2

X = 13,842.18

Therefore, after a year has passed, Peyton's account will be $ 13,842.18.

7 0
2 years ago
The range of a relation is
Kruka [31]
The range of a relation is the possible output values, the y-values.
3 0
3 years ago
Read 2 more answers
Factor <br> 5(x^2n-1)·(2x^3n 1)^2
nikdorinn [45]

I think I may be wrong check

5(x^2n-1)×(2x^3n-1) ^2

=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5

3 0
3 years ago
Steph makes 90 % 90%90, percent of the free throws she attempts. She is going to shoot 3 33 free throws. Assume that the results
zmey [24]

Using the binomial distribution, it is found that there is a 0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

For each free throw, there are only two possible outcomes, either he makes it, or he misses it. The results of free throws are independent from each other, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • He makes 90% of the free throws, hence p = 0.9.
  • He is going to shoot 3 free throws, hence n = 3.

The probability that he makes exactly 1 is P(X = 1), hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.9)^{1}.(0.1)^{2} = 0.027

0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

3 0
2 years ago
F(x)=-3x-5 and g(x) =4x-2 find (f-g)(x)
Triss [41]
<span>F(x)=-3x-5 and g(x) =4x-2 find (f-g)(x)

</span>(f-g)(x) = -3x-5 - (4x-2)
(f-g)(x) = -3x -5 -4x + 2
(f-g)(x) = -7x -3

hope that helps
3 0
3 years ago
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