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Furkat [3]
3 years ago
10

For each of the following series, tell whether or not you can apply the 3-condition test (i.e. the alternating series test). Ent

er D if the series diverges by this test, C if the series converges by this test, and N if you cannot apply this test (even if you know how the series behaves by some other test).1. ∑n=2 to [infinity]((−1)^n(n^4+2^n))/(n^3−1)2. ∑n=1 to [infinity](−1)^n/n^53. ∑n=1 to [infinity] ((−1)^ncos(n))/(n^2)4. ∑n=1 to [infinity] (((−1)^n(n^3+1))/(n^3+7)5. ∑n=1 to [infinity]((−1)^n(n^10+1))/e^n6. ∑n=1 to [infinity](((−1)^n(n^3+1))/(n^4+1)
Mathematics
1 answer:
Mila [183]3 years ago
6 0

Answer:

1. D  2. C  3. C  4. N 5. C  6. N

Step-by-step explanation:

The Alternating Series Test

∑(-1)nBn converges when the following two conditions are met:

(i) lim Bn = 0 and (ii) {Bn} is (eventually) decreasing.

Note: The Alternating Series Test doesn't apply when either of the conditions is not met, and so never is a test for divergence.  If the first condition isn't met, then the "n-th term test" will show divergence.

1. Bn = (n^4+2^n))/(n^3−1)

lim (n^4+2^n))/(n^3−1) = 0 but (ii) the function is increasing as n increases, Hence diverges

2. Bn = 1/n^5 lim Bn = 0. this is a converging p series as P>5 satisfies the condition for P series convergence P > 1 converges, 0<P<=1 diverges and (ii) is satisfied. Hence Converges

3. Bn = cos(n))/(n^2) ; cos(n) has no limit since it dangles between 0 and 1 and limit is unique. leaving 1/n^2 to converge according to the condition for P series convergence P > 1 converges, 0<P<=1 diverges and (ii) is satisfied. Hence Converges.

4. Bn = (n^3+1))/(n^3+7) ; lim Bn as n tends to infinity = 0. (ii) can be confirmed by limit compression test

5. Bn = (n^10+1))/e^n; lim Bn = 0 also Bn is a decreasing function since the denominator is an increasing function and (ii) is satisfied. Hence Converges.

6. Bn =(n^3+1))/(n^4+1) ; lim Bn as n tends to infinity = 0. (ii) can be confirmed by limit compression test

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In a survey of first graders, their mean height was 51.6 inches with a standard deviation of 3.6 inches. Assuming the heights ar
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Answer:

Step-by-step explanation:

In a survey of first graders, their mean height was 51.6 inches with a standard deviation of 3.6 inches. Assuming the heights are normally distributed, what height represents the first quartile of these students?

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2 years ago
10. 10. If AB bisects CAF, and m&lt;EAF = 72°,<br>then the m&lt;BAF =​
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The measure of ∠BAF is 54°.

Solution:

DF and CE are intersecting lines.

m∠EAF = 72° and AB bisects ∠CAF.

∠EAF and ∠DAC are vertically opposite angles.

Vertical angle theorem:

<em>If two lines are intersecting, then vertically opposite angles are congruent.</em>

∠DAC ≅ ∠EAF

m∠DAC = 72°

<em>Sum of the adjacent angles in a straight line = 180°</em>

m∠DAE + m∠EAF = 180°

m∠DAE + 72° = 180°

Subtract 72° from both sides.

m∠DAE = 108°

∠CAF and ∠DAE are vertically opposite angles.

⇒ m∠CAF = m∠DAE

⇒ m∠CAF = 108°

AB bisects ∠CAF means ∠CAB = ∠BAF

m∠CAB + m∠BAF = 108°

m∠BAF + m∠BAF = 108°

2 m∠BAF = 108°

Divide by 2 on both sides, we get

m∠BAF = 54°

Hence the measure of ∠BAF is 54°.

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3 years ago
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Answer:

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5 0
3 years ago
a. 0.98 – 0.053 b. 0.67 – 0.4 c. 0.3 – 0.002 d. 3.2 – .789 e. 6.53 – 4.298 f. 6 – 4.32 g. 7 – 3.574 h. 4.83 – 1.8 i. 3.7 – 1.8 j
Pavlova-9 [17]

a. 0.927

We have:

0.98 – 0.053

We can re-write it as:

0. 9 8 0 -

0. 0 5 3

Moving digits to the right:

0. 9 7 10 -

0. 0 5 3

Digit-per-digit subtraction:

0. 9 2 7


b. 0.27

We have:

0.67 – 0.4

We can re-write it as:

0. 6 7 -

0. 4 0

Digit-per-digit subtraction:

0. 2 7


c. 0.298

We have:

0.3 – 0.002

We can re-write it as:

0. 3 0 0 -

0. 0 0 2

Moving digits to the right:

0. 2 10 0 -

0. 0 0 2

Again:

0. 2 9 10 -

0. 0 0 2

Digit-per-digit subtraction:

0. 2 9 8


d. 2.411

We have:

3.2 – .789

We can re-write it as:

3. 2 0 0 -

0. 7 8 9

We need to rewrite the first term by moving digits to the right several times:

3. 2 0 0 = 2. 12 0 0 = 2. 11 10 0 = 2. 11 9 10

So now we have:

2. 11 9 10 -

0. 7 8 9

Digit-per-digit subtraction:

2. 4 1 1


e. 2.232

We have:

6.53 – 4.298

We can re-write it as:

6. 5 3 0 -

4. 2 9 8

We need to rewrite the first term by moving digits to the right several times:

6. 5 3 0 = 6. 5 2 10 = 6. 4 12 10

So now we have:

6. 4 12 10 -

4. 2 9 8

Digit-per-digit subtraction:

2. 2 3 2


f. 1.68

We have:

6 – 4.32

We can re-write it as:

6. 0 0 -

4. 3 2

We need to rewrite the first term by moving digits to the right several times:

6. 0 0 = 5. 10 0 = 5. 9 10

So now we have:

5. 9 10 -

4. 3 2

Digit-per-digit subtraction:

1. 6 8


g. 4.426

We have:

7 – 3.574

We can re-write it as:

7. 0 0 0 -

3. 5 7 4

We need to rewrite the first term by moving digits to the right several times:

7. 0 0 0 = 6. 10 0 0 = 6. 9 10 0 = 6. 9 9 10

So now we have:

6. 9 9 10 -

3. 5 7 4 =

Digit-per-digit subtraction:

3. 4 2 6


h. 3.03

We have:

4.83 – 1.8

We can re-write it as:

4. 8 3 -

1. 8 0

We can immediately do the digit-per-digit subtraction:

3. 0 3


i. 2.9

We have:

3.7 – 1.8

We can re-write it as:

3. 7 -

1. 8

We need to rewrite the first term by moving digits to the right:

3. 7 = 2. 17

So now we have:

2. 17 -

1. 8 =

Digit-per-digit subtraction:

2. 9


j. 4.538

We have:

16.17 – 11.632

We can re-write it as:

1 6 . 1 7 0 -

1 1 . 6 3 2

We need to rewrite the first term by moving digits to the right:

1 6. 1 7 0 = 1 6. 1 6 10 = 1 5. 11 6 10  

So now we have:

1 5. 11 6 10 -

1 1. 6 3 2 =

Digit-per-digit subtraction:

0 4. 5 3 8

3 0
3 years ago
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