For this case we have the following equation:
P (t) = P (1 + r / n) ^ (n * t)
Where,
P: initial investment
r: interest
n: periods
t: time
she will take on her 45th birthday:
for t = 25:
P (25) = 1000 * (1 + 0.0165 / 4) ^ (4 * 25)
P (25) = 1509.31 $
Answer:
The future value of this investment when she takes her trip is:
P (25) = 1509.31 $
Answer:
A score of 150.25 is necessary to reach the 75th percentile.
Step-by-step explanation:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.
This means that 
What score is necessary to reach the 75th percentile?
This is X when Z has a pvalue of 0.75, so X when Z = 0.675.




A score of 150.25 is necessary to reach the 75th percentile.
Just put the coefients in to a matrix
1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C-2%262%26-3%7C-14%5Cend%7Barray%7D%5Cright%5D%20)
mulstiply 2nd row by -1 and add to 3rd
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%262%260%7C-6%5Cend%7Barray%7D%5Cright%5D)
divde last row by 2
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 2rd row by 6 and add to top one
![\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 1st row by -1 and add to 2nd
![\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-3%260%260%7C6%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 2nd row by -3
![\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
mulstiply 2nd row by -1 and add to 1st row
![\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-3%7C-12%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 1st row by -3
![\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%7C4%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
rerange
![\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%7C-2%5C%5C0%261%260%7C-3%5C%5C0%260%261%7C%204%5Cend%7Barray%7D%5Cright%5D)
x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)
B is answer
Answer:
No.
First, multiply 6 x 4 ($4 is the amount of money he makes from each sale).
6 x 4 = $24
If Lin's goal is $32, and he sells 6 boxes of popcorn for only $24, then he will not reach his goal in that specified range.
F(3)=16 i saw this problem before so it should be the answer in the end