What is the answer to 15 divided by 9375
1 answer:
You might be interested in
Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right] \\\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
The slope m of a line
through points 
and 
is given by :<span>
Thus, the slope of the line passing through points </span><span>(−1, 7) and (2, 10) is
</span>
<span>
The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).
We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):
y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8
Answer: </span><span>C. −x+y=8</span>
Thus, the slope of the line passing through points </span><span>(−1, 7) and (2, 10) is
</span>
<span>
The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).
We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):
y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8
Answer: </span><span>C. −x+y=8</span>
Answer:
The numbers needed to fill each box are in the image attached
Step-by-step explanation:
The probability of the coin landing on heads is 2/7, so the probability of it landing on tails is 1 - 2/7 = 5/7
The probability of landing heads 2 times is:
P = (2/7) * (2/7) = 4/49
The probability of landing heads and then tails is:
P = (2/7) * (5/7) = 10/49
The probability of landing tails and then heads is:
P = (5/7) * (2/7) = 10/49
The probability of landing tails 2 times is:
P = (5/7) * (5/7) = 25/49
The numbers needed to fill each box are in the image attached.
