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Snezhnost [94]
3 years ago
9

If you have a stack of pennies without counting them how do you know if there is an even or odd number of them

Mathematics
1 answer:
zavuch27 [327]3 years ago
5 0

If I have a stack of pennies. And I have to tell that without counting whether there are even number of pennies or odd.

Even numbers are the numbers which are divisible by 2 and odd numbers are the numbers which are not divisible by 2.

Then I will put the given pennies in 2 rows and then I will match them to form a pair of 2 pennies. After matching, if there is 1 penny left over, then there is an odd number of pennies and if all the pennies have a match,then there is an even number of pennies.

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4. Assume that the chances of a basketball player hitting a 3-pointer shot is 0.4 and the probability of hitting a free-throw is
vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that p = 0.4. So this is P(X = 3) when n = 10.

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

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that p = 0.65. The probability is P(X = 5) when n = 10.

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

P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

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3 years ago
Simplify the expression.<br> The expression is<br> (5x+1/2)+(9x-2 1/2)
Alexxandr [17]
All you have to do here is combine like terms.
5x + 9x = 14x
.5 - 2.5 = -2

14x - 2
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