Answer:
$13,200
Step-by-step explanation:
You need to use the simple interest formula
I = P * r * t
I = Interest accrued
P = Principal amount invested
r = Interest rate you need to divide by 100 to get it in decimal form
t = time, in years if you are given a partial year, divide the months by 12
P = $12,000
r = 7.5% = .075
t = 1
But, because we want I to equal $990 then I is
I = $990
So we ignore our P and instead solve for the P that will give us the desired result.
I = P * r * t
$990 = P * .075 * 1
$990 = P.075 Divide each side by .075
$990/.075 = P.075/.075
$990/.075 = P
$13,200 = P
So, to earn an annual interest income of $990, $13,200 will have to be invested in the 7.5% bond.
Answer:
y = 18 / (x + 2)²
Step-by-step explanation:
Inversely proportional
y * (x + 2)² = k
when x=1, y=2
k = 2 * (1 + 2)²
k = 18
-----------------------
y * (x + 2)² = 18
Divide both sides by (x + 2)²
y = 18 / (x + 2)²
Using it's concept, the probability of choosing a number that is a multiple of 9 is of 0.111.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
From 1 to 81, there are 81 numbers, of which 81/9 = 9 are multiples of 9, hence, considering that 81 is the number of total outcomes and that 9 is the number of desired outcomes, the probability is given the following division:
p = 9/81 = 1/9 = 0.111.
More can be learned about probabilities at brainly.com/question/14398287
#SPJ1
Option D. D has the matrix of constants [[12], [11], [4]].
Step-by-step explanation:
Step 1:
With the given equations, we can form matrices to represent them.
The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.
Step 2:
The linear system A is represented as
![\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%261%261%5C%5C1%26-11%260%5C%5C1%26-1%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D26%5C%5C17%5C%5C23%5Cend%7Barray%7D%5Cright%5D)
.
Step 3:
The linear system B is represented as
![\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%261%261%5C%5C1%26-11%260%5C%5C1%26-1%2612%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D23%5C%5C17%5C%5C26%5Cend%7Barray%7D%5Cright%5D)
.
Step 4:
The linear system C is represented as
![\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C1%26-1%260%5C%5C1%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C11%5C%5C12%5Cend%7Barray%7D%5Cright%5D)
.
Step 5:
The linear system D is represented as
![= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%5C%5C11%5C%5C4%5Cend%7Barray%7D%5Cright%5D)
.
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
Answer:
lower 95% Cl is 0.292
Step-by-step explanation:
p^ = 0.4783
n = 23
for 95% Cl,
z = 1.9600
Inputting all the available data into P as attached below
LCL = 0.292
UCL = 0.670
therefore, lower 95% Cl = 0.292
