A(n)=ar^(n-1) so
100=4r^9
25=r^9
ln25=9lnr
ln25/9=lnr
r=e^(ln25/9)
a(n)=4e^((n-1)(ln25/9) if you wanted an approximation...
a(n)=4(1.43^(n-1))
Answer:
15
Step-by-step explanation: i'm not explaining why
If we divide 55 (the total amount of money he made) by 5 (price to walk one small dog), we get 11. Meaning that the maximum amount of small dogs he could possibly walk and receive that amount of money is 11. However, he only walked 8 dogs. Therefore Ryan would have needed to walk a specific amount of large dogs in order to earn 55 dollars in total.
After doing some process of elimination, I reached this conclusion:
3 small dogs = $15
5 large dogs = $40
The combination of the two would equal $55.
Therefore, the answer would be 3 small dogs.
2 cakes - Theresa's and Joe's
Theresa's cake had 6 pieces after she cut it. (2 times the size of Joe's pieces)
Joe's cake had 12 pieces after he cut it. (1/2 the size of Theresa's pieces)
We know that 8/12ths of ONE cake were eaten and that Joe ate 2 of his pieces.
We want to know how many pieces Theresa ate of her cake. Keeping in mind that her pieces are equal to 2 of Joe's pieces we can solve this question.
8/12 eaten total
if 2/12 by Joe
then 8-2 = 6, 6/12 by Theresa
(BUT: Theresa's pieces were twice the size of Joe's so we will divide by 2)
6/12 = 3/6
Answer: Theresa ate 3 pieces of her cake
Answer:
c. the sample portion is unbaised
Step-by-step explanation:
plato
