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3 years ago

What is the nth term to this sequence 18,16,14,12

Mathematics

2 answers:

densk [106]3 years ago

6 0

What is the nth term to this sequence 18,16,14,12 ?

Answer: 4

Explanation:

1. 18

2. 16

3. 14

4. 14

5. 12

6. 10

7. 8

8. 6

9. 4

Ratling [72]3 years ago

5 0

Answer:

Hey!

The nth term for your sequence is -2!

Step-by-step explanation:

18 - 2 = 16

16 - 2 =14

And so on! (the nth term of a sequence is basically the differences between the different terms!)

HOPE THIS HELPS!!

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4 years ago

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lina2011 [118]

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3 years ago

If i have three coins what is the possibility of getting at most three heads

sammy [17]

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4 years ago

Assume that all (525)poker hands are equally likely. What is the probability that you will be dealt:

AlekseyPX

Answer:

There is 1.98% of probability of being dealt a flush in 5-card Poker

Step-by-step explanation:

To know the probability of a flush being dealt, we can calculate the number of cases when that happens and divide it by the total number of cases of poker hands that exist, naming A the event of a flush.

We will use combinations (nCr button on a calculator) to count the number of cases, because we don't care about the order (it is the same to be dealt a 2, 4, 6, 7 and 8 of hearts than the opposite order), being a flush the event when we take 5 cards out of 13 of the same suit, times 1 out of 4 possible suits and the total number of cases is taking 5 random cards out of 52.

$P_{A} =\frac{Cases Of Flush}{Hands} =\frac{\left(\begin{array}{ccc}13\\5\end{array}\right)*\left(\begin{array}{ccc}4\\1\end{array}\right)}{\left(\begin{array}{ccc}52\\5\end{array}\right)} =\frac{5148}{2598960}=0,00198\\$

That means there is about a 2% of probability of being dealt a flush.

In other words, of every 16660 plays, 33 will be, on average, a flush

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3 years ago

Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }$

3 0

3 years ago