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ruslelena [56]
3 years ago
11

In a test for extrasensory perception, the experimenter looks at cards containing either a star, circle, wave, or square. (The s

ubject cannot see the cards.) As the experimenter looks at each of 20 cards, the subject names the shape on the card. Assuming that any success guessing shapes is due purely to chance, what is the probability a subject correctly guesses at least 10 of the 20 shapes? (Round to 3 decimal places)
Mathematics
1 answer:
babymother [125]3 years ago
4 0

Answer:

The probability of a subject correctly guesses at least 10 of the 20 shapes is 0.014

Step-by-step explanation:

Looking at the experiment, we can find the probability thinking it as a Bernoulli experiment (dicotomic variable), as the subject either guess the shape right or not, and each guess is independent from the others (as are due to chance).

We are going to use a binomial distribution (useful to model successes in n   Bernoulli experiments), as we want to know the probability of at least 10 right guesses (number of sucess) in 20 cards (number of independent experiments). The probability of guessing right a shape is:

\mbox{Probability of guessing right}=\frac{\mbox{favorable cases}}{\mbox{total cases}}=\frac{1}{4}

The probability of k sucesses in n experiments, each with a p probability of sucess is given by:

P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}

As we want to know the probability of getting at least 10 correct guesses, we have to add the odds for 10, 11, 12, ..., 20 right guesses:

P(X \ge 10) = \sum_{i=10}^{20} {20\choose i}0.25^i(1-0.25)^{20-i}=0.014

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5/7 = 150 / ?

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5 * ? = 7 * 150

? = 7 * 150 / 5

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B. The Slope is 0




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135°<br> Solve for &lt;2.<br> 62 = [?]<br> 45°<br> &amp;2<br> 86°<br> Enter
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\displaystyle  \angle2     =  {49}^{ \circ}

Step-by-step explanation:

remember that,

<u>if a side of a triangle is extended then exterior angle so formed equal to the sum of the two opposite interior </u><u>angles</u>

thus our equation is

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A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market. Which amount is closest to
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2.33 kg

Step-by-step explanation:

If the baker needs to use 2.327 kilograms of blueberries, we are going to subtract the amount of the packages to the amount he needs and we'll see which is the smallest number.

In the case of 2.3, 2.327 - 2.3 = .027

In the case of 2.42, 2.327 - 2.42 = -0.09

In the case of 2.33, 2.327 - 2.33 = -0.003

In the case of 2.4, 2.327 - 2.4 = -0.07

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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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