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3 years ago

What is the common ratio for the sequence 5,-10,20,-40?

Mathematics

1 answer:

algol [13]3 years ago

8 0

Answer:

The COMMON RATIO of the sequence is -2.

Step-by-step explanation:

Here, the given sequence is:

5, -10, 20 ,-40

Here, The first term ( a 1 ) of the sequence is (5)

The second term (a 2) of the sequence is (-10)

Third term of sequence (a 3) is 20

Now, calculating the common ratio (r) of sequence ,we get:

$\frac{a_2}{a_1} = r = (\frac{-10}{5}) = -2\\\frac{a_3}{a_2} =r = (\frac{20}{-10}) = -2$

So, here r =(-2)

⇒ The COMMON RATIO of the sequence is -2.

Hence, the given sequence in in GEOMETRIC PROGRESSION.

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3 years ago

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rosijanka [135]

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4 years ago

What is the simplified form of the expression 3 + 8x − 2y − x + 1 + 4y

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3 years ago

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The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associate

Schach [20]

Answer:

(a)0.16

(b)0.588

Step-by-step explanation:

The matrix below shows the transition probabilities of the state of the system.

$\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix $P^k$ .

(a)

$P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

$P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}$

If the system is initially running, the probability of the system being down in the next hour of operation is the $(a_{12})th$ entry of the P^2$ matrix.$

The probability of the system being down in the next hour of operation = 0.16

(b)After two(periods) hours, the transition matrix is:

$P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}$

Therefore, the probability that a system initially in the down-state is running

is 0.588.

(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.

Since we have two states, $S=[s_1$ s_2]$

$[s_1$ s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$ s_2]$

Using a calculator to raise matrix P to large numbers, we find that the value of $P^k$ approaches [0.75 0.25]:

Furthermore,

$[0.75$ 0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75$ 0.25]$

The steady-state probabilities of the system being in the running state and in the down-state is therefore:

$[s_1$ s_2]=[0.75$ 0.25]$

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4 years ago

Arlan needs to create a box from a piece of cardboard. The dimensions of his cardboard are 10 inches by 8 inches. He must cut a

raketka [301]

1) Dimensiones of the cardboard:

length: 10 inches
width: 8 inches

2) dimensions of the squares cut

length: x
width: x

3) dimensions of the box:

length of the base = 10 - 2x
width of the base = 8 - 2x
height = x

4) Volume of the box

V = (10 - 2x) (8 - 2x) x = x [80 - 20x - 16x + 4x^2] = x [ 80 - 36x + 4x^2 ] =

V = 80x - 36x^2 + 4x^3

5) Maximum volume => derivative of V, V' = 0

V' = 80 - 72x + 12x^2 = 0

6) Solve the equation

Divide by 4 => 3x^2 - 18x + 20 = 0

Use the quadratic formula: x = 1.47 and x = 4.53 (this is not valid)

So, the answer is the option C. 1.5 inches.

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4 years ago