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2 years ago

What is the maximum percent of net spendable income that should be set aside for housing

Mathematics

2 answers:

LUCKY_DIMON [66]2 years ago

8 0

PLATO ANSWERS:

Transportation= 20%

Housing= 36%

Food= 17%

Hope I helped, have a great day!

stepladder [879]2 years ago

7 0

Answer:

Answer;

-38 %

Step-by-step explanation:

Explanation;

-Budget busters are the large potential problem areas that can destroy a budget. Failure to control even one of these problem areas can result in financial disaster.

-Housing takes about 38 percent of your monthly budget. Housing decisions should be based on need and financial ability, not on internal or external pressure.

-Food takes 12 percent of your monthly budget. The reduction of a family's food bill requires quantity and quality planning.

-Transportation (purchase and maintenance), takes 15 percent of your monthly budget, Debts takes 5 percent of Net Spendable Income, Insurance takes 5 percent of Net Spendable Income assuming an employer provides medical insurance, Recreation/Entertainment takes 5 percent of Net Spendable Income, Clothing takes 5 percent of Net Spendable Income, Medical and dental takes 5 percent of Net Spendable Income and Savings takes 5 percent of Net Spendable Income

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The cost of a car is $15,570. You plan to make a down payment of $1,500, and a monthly payment of $338.08 for 60 months. What is

Helga [31]

Hi there
1) b
15,570−1,500
=14,070
2) a
338.08×60
=20,284.8
3)d
20,284.8−14,070
=6,214.8
4) c
338.08×60+1,500
=21,784.8

Hope it helps

4 0

2 years ago

Read 2 more answers

It is known that a cable with a cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the

vesna_86 [32]

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

5 0

2 years ago

Read 2 more answers

Select the correct answer.

igor_vitrenko [27]

Answer:

Step-by-step explanation:

CPI stands for Consumer Price Index, which is the measure of average prices to goods and services.

So, in this case, we have to find the average price for the given basket in year 2.

We know that in year 2, a toothbrush tube costs $1.25 and a tube of toothpaste costs $2.10. If we have 5 toothbrushes and 2 tubes of toothpaste, the total cost would be

$5 \ toothbrushes=5(\$1.25)=\$6.25\\2 \ toothpaste \ tubes=2(\$2.10)=\$4.20$

Now, the CPI would be the sum of these costs

$CPI = \$6.25 + \$4.20=\$10.45$

Therefore, the Consume Price Index for this basket of goods is

B. $\$10.45$

7 0

2 years ago

Read 2 more answers

Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe

sukhopar [10]

Answer:

$x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the $\bar x$ so that $A \bar x = 0$

$\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]$

Now the augmented matrix is

$\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]$

After this, we decrease this to reduce the formation of the row echelon

$R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]$

$R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]$

$R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0$

$x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

By applying the above matrix, we can easily reach an answer

5 0

2 years ago

4.52 Random variables X and Y follow a joint distribution f(x, y) = 2, 0 < x ≤ y < 1, 0, otherwise. Determine the correlat

VMariaS [17]

First, you'll need to find the marginal distributions of $X,Y$ . By the law of total probability,

$P(X=x)=\displaystyle\sum_yP(X=x)P(Y=y)$

which translates to

$f_X(x)=\displaystyle\int_x^1f_{X,Y}(x,y)\,\mathrm dy=\begin{cases}2(1-x)&\text{for }0$

Similarly,

$f_Y(y)=\displaystyle\int_0^yf_{X,Y}(x,y)\,\mathrm dx=\begin{cases}2y&\text{for }0$

Compute the expectations for both random variables:

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13$

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23$

Compute the variances and thus standard deviations:

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16$

$\implies V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}\implies\sqrt{V[X]}=\dfrac1{3\sqrt2}$

$E[Y^2]=\displaystyle\int_{\infty}^\infty y^2f_Y(y)\,\mathrm dy=\int_0^12y^3\,\mathrm dy=\frac12$

$\implies V[Y]=\dfrac12-\left(\dfrac23\right)^2=\dfrac1{18}\implies\sqrt{V[Y]}=\dfrac1{3\sqrt2}$

Compute the covariance:

$\operatorname{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We have

$E[XY]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_0^y2xy\,\mathrm dx\,\mathrm dy=\frac14$

and so

$\operatorname{Cov}[X,Y]=\dfrac14-\dfrac13\dfrac23=\dfrac1{36}$

Finally, the correlation:

$\operatorname{Corr}[X,Y]=\dfrac{\operatorname{Cov}[X,Y]}{\sqrt{V[X]}\sqrt{V[Y]}}=\dfrac{\frac1{36}}{\left(\frac1{3\sqrt2}\right)^2}=\dfrac12$

3 0

2 years ago