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2 years ago

Find the first five terms of the sequence given by a 1 = 2, a n = 3 a n -1 - 1.

Mathematics

2 answers:

lara31 [8.8K]2 years ago

8 0

Answer:

2, 5, 14, 41, 122

Step-by-step explanation:

Using the recursive rule with a₁ = 2

a₂ = 3a₁ - 1 = 3(2) - 1 = 6 - 1 = 5

a₃ = 3a₂ - 1 = 3(5) - 1 = 15 - 1 = 14

a₄ = 3a₃ - 1 = 3(14) - 1 = 42 - 1 = 41

a₅ = 3a₄ - 1 = 3(41) - 1 = 123 - 1 = 122

The first 5 terms are 2, 5, 14, 41, 122

Butoxors [25]2 years ago

7 0

Answer:

$a_{1}\\\\$ = 2

$a_{2\\}\\\\$ = 5

$a_{3\\}\\\\$ = 14

$a_{4\\}\\\\$ = 41

$a_{5\\}\\\\$ = 122

Step-by-step explanation:

If I've properly understood:

$a_{1}\\\\$ = 2

$a_{n} = 3 a_{n-1}\\$ - 1

so, the first 5 terms are:

$a_{1}\\\\$ = 2

$a_{2\\}\\\\$ = 3. 2 - 1 = 5

$a_{3\\}\\\\$ = 3. 5 - 1 = 14

$a_{4\\}\\\\$ = 3. 14 - 1 = 41

$a_{5\\}\\\\$ = 3. 41 - 1 = 122

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The Edward James Toy Company uses a Kanban system to make plastic wheels that are a component of several toys. The waiting time

pickupchik [31]

Answer:

$N_c \approx 5.0$

Step-by-step explanation:

From the question we are told that:

Waiting time for a container of the wheels during production $T_w=0.25day$

Average processing time is $T_a=0.15 day per container$

Wheel per container $N_w=200$

Total wheel per day $D=2000$

Policy variable of $X=5%$

Generally the equation for Total number of container $N_c$ is mathematically given by

$N_c=DT(\frac{1+X}{N_w} )$

Where

Total time $T=T_w+T_a$

$T=0.15+0.25$

Therefore

$N_c=DT(\frac{1+X}{N_w} )$

$N_c=2000*0.40(\frac{1+0.5}{200} )$

$N_c=(\frac{840}{200} )$

$N_c=4.2$

$N_c \approx 5.0$

Therefore the number of Kanban containers needed for the wheels is

$N_c \approx 5.0$

3 0

2 years ago

The length of a shoestring is 21 1/4 centimeters long. What would be the length of 4 of these shoe strings?

beks73 [17]

Answer:

21.25 multiplied by 4, which is 85. Final Answer : 85 centimeters long

Step-by-step explanation:

21.25 is the same as 21 1/4.

3 0

2 years ago

Read 2 more answers

What is Permutations and Combinations?

marissa [1.9K]

Answer:

See below

Step-by-step explanation:

Permutation is to select an object then arrange it and it cares about the orders while Combination is about only selecting an object without caring the orders.

Permutation can be expressed in math as:

$\displaystyle{_n P _r = \dfrac{n!}{(n-r)!} \ \ \ (n \geq r) }$

where n is a number of total object and r is a number of selected object to arrange. Hence. n cannot be less than r.

Now let's see an example of permutation, suppose we have letter A, B and C. I'd like to know how many ways these words can be arranged:

Since there are 3 letters total and 3 selected letters to arrange then:

$\displaystyle{_3 P _3 = \dfrac{3!}{(3-3)!}}\\\\\displaystyle{_3 P _3 = \dfrac{3 \times 2 \times 1}{0!}}\\\\\displaystyle{_3 P _3 = \dfrac{6}{1}}\\\\\displaystyle{_3 P _3 = 6}$

Therefore, there are 6 ways to arrange the letters - we can also demonstrate visually:

ABC - 1

ACB - 2

BAC - 3

BCA - 4

CAB - 5

CBA - 6

Notice that if you do visually, you'll get the same answer as the calculation of permutation!

----

Combination can be expressed mathematically as:

$\displaystyle{_n C _r = \dfrac{n!}{(n-r)!r!} = \dfrac{_n P _r}{r!} \ \ \ (n \geq r) }$

The difference between permutation and combination is that you only find how many ways you can select object in combination. Therefore, no arrange and doesn't care about order, just ways to select.

Suppose we have same 3 letters: A, B and C. I want to find how many ways I can select these 3 letters:

Since there are 3 letters total and 3 selected letters:

$\displaystyle{_3 C _3 = \dfrac{3!}{(3-3)!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{0!3!}}\\\\\displaystyle{_3 C _3 = \dfrac{3!}{3!}}\\\\\displaystyle{_3 C _3 = 1}$

Hence, there is only one way to select 3 letters. This makes sense because if you have 3 letters then you can only select 3 letters only one way.

5 0

2 years ago

How does the Moon reflect sunlight as it orbits Earth, creating predictable<br>cyclic patterns?

kogti [31]

Hmmm very good question

4 0

2 years ago

PLEASE HELP ME

denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

(2, 1), (2, 2), (2, 3), (2, 4)

(3, 1), (3, 2), (3, 3), (3, 4)

(4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

$P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

$E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36$

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

$P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

$E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24$

Thus, the number of times should Hartley spin a difference of 0 is 24.

4 0

2 years ago