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3 years ago

State the common difference for the given A.P.3/2,1/2,-1/2,-3/2,......

Mathematics

1 answer:

a_sh-v [17]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

The common difference is the constant amount of change between the numbers in an arithmetic sequence.

To find the common difference, you take the last term and subtract it by the previous term.

$\frac{-3}{2} - \frac{-1}{2} =-1$

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[|(x-4)÷(x+5)|]\leq[4]

Katena32 [7]

Answer:

Solutions are;

x = -8

x = -3.2

Step-by-step explanation:

Here, we want to solve the given equation for x

|(x-4)/(x + 5)| = 4

From what we have, this is an absolute value equation and thus, we are going to have two solutions

These are;

x-4/x+ 5 = 4

x-4 = 4(x + 5)

x-4 = 4x + 20

x-4x = 20 + 4

-3x = 24

x = -24/3

x = -8

secondly;

x-4/x+5 = -4

x-4 = -4(x + 5)

x-4 = -4x - 20

x + 4x = -20 + 4

5x = -16

x = -16/5

x = -3.2

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3 years ago

Prompt Use the applet to select 20 different random samples of 100 part-time college students. For each sample the applet calcul

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Answer:

Yes, we can expect the given result

Step-by-step explanation:

The population proportion P = 0.60

the 95% confidence interval is ( 0.56, 0.76)

The 95% confidence that the true population proportion lies between the interval ( 0.56, 0.76)

Only 5% of 20 random samples i.e only 1 sample is not contain in the population proportion.

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3 years ago

URGENT!!!!! What is the value is x? A. 5.375 B. 5.5 C. 11 D. 13

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Answer:

value of x is B 5.5

Step-by-step explanation:

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Answer:

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

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3 years ago

State the common difference for the given A.P.3/2,1/2,-1/2,-3/2,......​

State the common difference for the given A.P.3/2,1/2,-1/2,-3/2,......