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1 year ago

) Studies also find that teenagers have a higher price elasticity than do adults. Why might this be true?

Business

1 answer:

Andru [333]1 year ago

5 0

This statement is true. Teenagers have a higher price elasticity than do adults.

This is because teenagers does not have much income to spend so they can shift to other products also. Teenagers are also not addict to smoking or any other habits , Therefore price affects the demand of the product.

Price elasticity is defined as the ratio of the percentage change in quantity demanded due to percentage change in price.

Elasticity having more than 1 are relatively more elastic.

Adults have more income to spent so the demand do not effects due to change in price. As adults are more addicted to smoking then teenagers ,

addicted people did not consume less due to its price change.

To know more about price elasticity:

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3 years ago

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3 years ago

A bag contains 9 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find t

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a) 0.0260

b) 0.0681

Explanation:

The bag contains:

r = 9 (number of red marbles)

w = 5 (number of white marbles)

b = 6 (number of blue marbles)

So, the total number of marbles in the bag at the beginning is:

$n=9+5+6=20$

At the 1st attempt, the probability of choosing a red marble is:

$p(r)=\frac{r}{n}=\frac{9}{20}$ (1)

At the 2nd attempt, the 1st red marble is not placed back, so now the number of marbles is (n-1), while the number of red marbles left is (r-1). So the probability of choosing another red marble at the 2nd attempt is

$p(r)=\frac{r-1}{n-1}=\frac{8}{19}$ (2)

With a similar argument, the probabilities of selecting a red marble in the 3rd and 4th attempt are

$p(r)=\frac{7}{18}$ (3)

$p(r)=\frac{6}{17}$ (4)

Therefore, the probability of drawing 4 red marbles in the first 4 attempts without replacing is:

$p(rrrr)=\frac{9}{20}\frac{8}{19}\frac{7}{18}\frac{6}{17}=\frac{3024}{116280}=0.0260$

At the 1st draw, the probability that the marble is not red is:

$p(r^c)=1-p(r)=1-\frac{9}{20}=\frac{11}{20}$

At the 2nd draw, there are 9 red marbles left and 19 total marble left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{19}=\frac{10}{19}$

At the 3rd draw, there are 9 red marbles left and 18 total marbles left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{18}=\frac{9}{18}$

Finally, with a similar argument the probability of NOT drawing a red marble at the 4th and last attempt is:

$p(r^c)=\frac{8}{17}$

So, the total probability of drawing 4 non-red marbles in the first 4 attempts is:

$p(r^cr^cr^cr^c)=\frac{11}{20}\frac{10}{19}\frac{9}{18}\frac{8}{17}=\frac{7920}{116,280}=0.0681$

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Answer:

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