Answer:
hola yo tambien tengo 13
Step-by-step explanation:
When n = 1, 2(1) + 1 = 3
When n = 2, 2(2) + 1 = 5
When n = 3, 2(3) + 1 = 7
When n = 4, 2(4) + 1 = 9
When n = 5, 2(5) + 1 = 11
Sum = 3 + 5 + 7 + 9 + 11 = 35
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Answer: 35
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well, we're assuming all along that Merina owes Bradford $2000, because in the 1st scenario, she was going to pay twice $1000.
on the 2nd scenario, she'll be paying the same $2000 but split 7 months from now and then 7 months later, same 2000 bucks, at which point Bradford applied 8.5% interest.
using those assumptions, since the wording is not quite clear, we can say that Merina is simply paying 2000 bucks plus the 8.5%
![\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8.5\% of 2000}}{\left( \cfrac{8.5}{100} \right)2000}\implies 170 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\stackrel{principal}{2000}~~ + ~~\stackrel{interest}{170}}{2}\implies \stackrel{\textit{two equal payments of}}{1085}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Ba%5C%25%20of%20b%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20%5Cleft%28%20%5Ccfrac%7Ba%7D%7B100%7D%20%5Cright%29%5Ccdot%20b%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7B8.5%5C%25%20of%202000%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B8.5%7D%7B100%7D%20%5Cright%292000%7D%5Cimplies%20170%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%5Cstackrel%7Bprincipal%7D%7B2000%7D~~%20%2B%20~~%5Cstackrel%7Binterest%7D%7B170%7D%7D%7B2%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Btwo%20equal%20payments%20of%7D%7D%7B1085%7D)
Answer:
We are 95% sure that the true proportion of students that supports a fee increase is between 0.75 and 0.85.
Step-by-step explanation:
The interpretation of a x% confidence interval of proportions being between a and b is that:
We are x% sure that the true proportion of the population is between a and b.
If the 95% confidence interval estimating the proportion of students supporting the fee increase is (0.75; 0.85), what conclusion can be drawn
We are 95% sure that the true proportion of students that supports a fee increase is between 0.75 and 0.85.
Answer:
You can solve it by saying half times base times perpendicular high
