(a) 
since 13 is prime.
(b) 
, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so 
.
(c) 
; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with 
.
(d) 
; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so 
.
The Mixture B is the answer! Hope this helps
Answer: i - j - k
Step-by-step explanation:
Taking the cross product between two vectors will give you a third vector that is orthogonal(perpendicular) to both vectors.
<1,1,0> x <1,0,1>
![det(\left[\begin{array}{ccc}i&j&k\\1&1&0\\1&0&1\end{array}\right] )](https://tex.z-dn.net/?f=det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%260%5C%5C1%260%261%5Cend%7Barray%7D%5Cright%5D%20%29)
the determinate of the matrix: <1,-(1),-1>
or: i - j - k
Answer:
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 27 - 1 = 26
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 2.7787.
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.
The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.
The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).
950 * .05 = $47.5
550 * .04 = $22
$47.5 + $22 = $69.50 interest earned
answer: amount in 4% account = $550
