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murzikaleks [220]

3 years ago

9

Select three common forms of Anglo-Saxon literature. lyric novella riddle epic elegy ballad novel short story

2 answers:

Solnce55 [7]3 years ago

8 0

Epic, elegy, and ballad are three common forms of Anglo-Saxon literature, whereas lyric, novella, riddle, novel, and short story were created much later.

Kipish [7]3 years ago

3 0

Answer:

The three common forms of Anglo-Saxon literature are:

<u>Epic, Ballad and Elegy.</u>

Step-by-step explanation:

Anglo-Saxon literature is covered since the fifth century in the Anglo-Saxon (Old English) period of Great Britain.

This literature covers genres as diverson as the epic, the ballad, the elegy, translations of the Bible and sermons.

-The epic is a genre that shows us legendary or fictional deeds of heroic feats whether real or imaginary in which they participated. An example is <em>Homer's Odyssey.</em>

-A ballad is a type of romantic song where musical tones are created from the writing of poems. Telling a person's life or some precise facts.

-The elegies are poems written in the first person, implying the ephemeral nature of joy, wealth and luck in life. These are works that fall into the sadness of the past. An example is <em>the elegy of The Wandering Man</em>.

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Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .

Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b) "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e) "=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f) "=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got $t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228$

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have $\alpha/2=0.025$ .

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got $t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086$

c. Central area =.99, df = 20

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got $t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845$

d. Central area =.99, df = 50

For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have $\alpha/2=0.005$ .

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got $t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678$

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means $\alpha =0.01$

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got $t_{\alpha}= 2.485$

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means $\alpha =0.025$

We can use the following excel code:

"=T.INV(0.025,5)"

And we got $t_{\alpha}= -2.571$

8 0

3 years ago

PLEASE HELP ITS DUE IN 10 MINUTES!!!!!

jekas [21]

Answer:

76 Square Feet

Step-by-step explanation:

First, add all the numbers you see up. Your final answer of square feet will be the total you added up as a whole.

24+14+4+6+5+3+15+5= 76

6 0

2 years ago

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Anyone ??<br> I’m confused

Ad libitum [116K]

+2+(-3)+(-1)+4(-5)+(-3) ok so first 2 is positive and 3 is negative there both different signs so u subtract +2+(-3)=-1+(-1)=0+4=4+(-5)=-1+(-3)=-4 your answer is 4

4 0

3 years ago

Please Help ASAP!!!!!!!!!

Ghella [55]

Answer:

Use math

Algebra calculator

Step-by-step explanation:

6 0

3 years ago

Solve for x:<br> 7x = 105

ohaa [14]

Answer:

The answer is 15 !!!

Step-by-step explanation:

Divide 105 by 7 and get 15. The answer to your question is x=15.

5 0

3 years ago

Read 2 more answers