Ask question

Ask question

All categories

Oksi-84 [34.3K]

2 years ago

7

You have a realized income of $3,264.71/month. Your housing and fixed expenses are 36% of your monthly realized income. You want

to save 4 months of an emergency fund within 9 months. How much must you save each month?

1 answer:

OLEGan [10]2 years ago

3 0

Answer:

$522.36

Step-by-step explanation:

First we need to find out how much money 4 months of an emergency fund is total.

36% of our monthly income is $1,175.30.

Our total emergency fund will be:

$1,175.30 x 4 months = $4,701.2

Now that we have a value for our total emergency fund, we then divide the total amount by 9.

$4,701.2/9 = $522.36

We would need to save $522.36 each month for 9 months.

You might be interested in

A 30 -60 -90 degree triangle has a hypotenuse with a length of 10. What is the length of the longer leg of the triangle?

Degger [83]

Answer:

the right answer is 5√3 believe me i took this quiz

Step-by-step explanation:

6 0

3 years ago

3 and 1/13 in simplest form.

defon

Answer:

3 1/3 is already in the simplest form. It can be written as 0.230769 in decimal form (rounded to 6 decimal places).

Have an awesome day! :D

8 0

2 years ago

Read 2 more answers

You need to establish _____ for how to judge a concept.

makvit [3.9K]

Answer:

Criteria.

Step-by-step explanation:

A series of criteria is a set of parameters established by whoever seeks to interpret a certain information or quantity of concepts, by means of which a set of interpretive rules is established that should be equal to all the concepts to be analyzed. Thus, the criteria allow a set of data to be analyzed under the same parameters or information processing requirements.

6 0

3 years ago

The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar

Alexus [3.1K]

Question:

The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar.

Given:

Cost of 12 kg sugar = $240

To find?

cost of 3 kg sugar

Answer :

<u>To find </u><u>c</u><u>o</u><u>s</u><u>t</u><u> </u><u>of</u><u> </u><u>3kg sugar first we have to find cost of 1 kg sugar</u><u>.</u>

Cost of 1 kg sugar = Cost of total no. of sugar ÷Total sugar
Cost of 1 kg sugar = $240/12
Cost of 1 kg sugar = $20

<u>Now Let's find cost of 3 kg sugar</u>

Cost of 3kg sugar = total no of sugar ×Cost of 1 kg sugar

Cost of 3kg sugar = 3×$20

Cost of 3kg sugar = $60

3 0

2 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago