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Leya [2.2K]
2 years ago
8

How many different sequences of 3 playing cards (from a single 52-card deck) exist?

Mathematics
1 answer:
hichkok12 [17]2 years ago
4 0

There 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

<h3>What is Permutations?</h3>

A permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.

Here, total number of cards = 52

         Number of Withdrawn cards = 3

Possibilities of different sequences of 3 playing cards = 52 X 51 X 50

                                                                                          = 132600

Thus, there 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

Learn more about Permutations from:

brainly.com/question/1216161

#SPJ1

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Answer:

Add the two in the middle and multiply it by the outside number

Step-by-step explanation:

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3 years ago
Line l contains points (0, -2) and (6, 6). Point P has coordinates (-1, 5)
barxatty [35]
Whats the question because I can probably help you
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The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.
Anton [14]

Answer:

a) The evidence suggest the true mean life of a biomedical device > 5500

b) (5487.94, +∞)

c) There is evidence to support that the mean is equal to 5500

Step-by-step explanation:

Here we have

(a) To test the hypothesis we have the claim that the mean life of biomedical devices > 5500

Therefore, we put the null Hypothesis which is the proposition that a difference does not exist. That is

H₀:  μ = 5500

Therefore, the alternative hypothesis will be

Hₐ:  μ > 5500

The test statistic is then found by;

t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}} =\frac{5617.8-5500 }{\frac{234.5 }{\sqrt{15}}} \approx 1.95

The P value from the T table at df = n - 1 = 15 - 1 = 14 is

0.025 < P < 0.05

The P value is given as 0.036 from the T distribution at 14 derees of freedom df

Therefore, since P < α or 0.05, we reject the null hypothesis. That is we fail to reject Hₐ:  μ > 5500. The evidence suggest the true mean life of a biomedical device > 5500

(b) The 95% confidence interval is given as

CI=\bar{x}\pm t_\alpha \frac{s}{\sqrt{n}}

Which gives    t_\alpha =  \pm2.145

CI=5617.8\pm 2.145 \times \frac{234.5}{\sqrt{15}}

The confidence interval of the lower bound on the mean is then

(5487.94, +∞)

c) From the above result, we find that the mean of 5500 is contained in the range of the lower confidence on the mean. We can therefore, accept the null hypothesis. That is there is evidence to support that the mean is equal to 5500.

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3 years ago
How to do linear functions using the table ?
Ira Lisetskai [31]

Answer:

-8 = m-2+b

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4 = m2+b

10 = m4+b

Step-by-step explanation:

You just put them in y = mx+b format

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