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enot [183]
3 years ago
7

Show work explain with formulas.

Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

1. a

44

=481,

n=1

∑

∞

(11n−3), the sum not exist

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

3.

n=1

∑

47

(9n+16)=10,904

\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}

4.

n=1

∑

∞

(6(2

n

)), the sum not exist

Step-by-step explanation:

\begin{lgathered}1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481\end{lgathered}

1.

8+19+30+41+...

19−8=11

30−19=11

41−30=11

It’s an arithmetic series. a

1

=8, d=11.

The formula for the n-th term of an arithmetic sequence:

a

n

=a

1

+(n−1)d

Substitute:

a

n

=8+(n−1)(11)use the distributive property a(b−c)=ab−ac

a

n

=8+11n−11

a

n

=11n−3

Calculate the 44th term. Put n = 44 to the formula:

a

44

=(11)(44)−3=484−3=481

\begin{lgathered}\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.\end{lgathered}

The summation notation:

n=1

∑

∞

(11n−3)

The sum not exist, because d>1, therefore 11n−3→∞.

=======================================================

\begin{lgathered}3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47\end{lgathered}

3.

25+34+43+52+...+436

34−25=9

43−34=9

52−43=9

It’s an arithmetic series. a

1

=25, d=9.

The formula for the n-th term of an arithmetic sequence:

a

n

=a

1

+(n−1)d

Substitute:

a

n

=25+(n−1)(9)

a

n

=25+9n−9

a

n

=9n+16

Calculate which the term of arithmetic series is the number 439.

Put a

n

=439:

9n+16=439subtract 16 from both sides

9n=423divide both sides by 9

n=47

\begin{lgathered}\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904\end{lgathered}

Therefore we have the series in summation notation:

n=1

∑

47

(9n+16)

For calculation of the sum we use the formula of a sum of terms

of an arithmetic sequence:

S

n

=

2

a

1

+a

n

⋅n

Substitute n=47, a

1

=25, a

47

=439:

S

47

=

2

25+439

⋅47=

2

464

⋅47=232⋅47=10904

n=1

∑

47

(9n+16)=10,904

=======================================================

\begin{lgathered}4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty\end{lgathered}

4.

12+24+48+...

24:12=2

48:24=2

It’s a geometric series a

1

=12, r=2.

The formula for the n-th term of a geometic sequence:

a

n

=a

1

r

n−1

Substitute:

a

n

=(12)(2

n−1

)use

a

m

a

n

=a

n−m

a

n

=(12)(

2

1

2

n

)

a

n

=(6)(2

n

)

The summation notation:

n=1

∑

∞

(6(2

n

))

The sum not exist, because r>1, therefore 6(2

n

)→∞

Sedaia [141]3 years ago
6 0

Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}

Step-by-step explanation:

1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481

\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.

=======================================================

3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47

\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904

=======================================================

4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty

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The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes
Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z=  X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

​

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute  how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z=  X−μ /σ

Z=90−77 /12

Z​=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

​

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

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Step-by-step explanation:

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