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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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crimeas [40]

Answer:

coffee is the best answer in my mind

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2 years ago

Mrs. Kong uses 1 over 3 of a stick of butter in a sauce she then uses 5 over 8 of the butter to make garlic bread what fraction

Allisa [31]

Answer=1/24

1/3+5/8=
to solve for this, we need the denominators of the fractions to match.

The LCM of 3 and 8 is 24

1/3=8/24
5/8=15/24

8/24+15/24=23/24

Now if we're using 23/24, then only 1/24 is left.

24/24-23/24=1/24

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1 year ago

Mark has three 4 1/2 oz cans and five 8 1/4 oz cans. How many ounces of tomatoes does Mark need?

Mashutka [201]

Answer:

54.74 ounces of tomatoes mark need.

Step-by-step explanation:

Given : Mark has three $4\frac{1}{2}$ oz cans and five $8\frac{1}{4}$ oz cans.

To find : How many ounces of tomatoes does Mark need?

Solution :

Mark has three $4\frac{1}{2}$ oz cans.

i.e. $3\times 4\frac{1}{2}=3\times \frac{9}{2}=\frac{27}{2}$

Mark has five $8\frac{1}{4}$ oz cans.

i.e. $5\times 8\frac{1}{4}=5\times \frac{33}{4}=\frac{165}{4}$

Total ounces of tomatoes does Mark need is

$T=\frac{27}{2}+\frac{165}{4}$

$T=\frac{54+165}{4}$

$T=\frac{219}{4}$

$T=54.75$

Therefore, 54.74 ounces of tomatoes mark need.

7 0

1 year ago

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

1 year ago

Given: △ABC, m∠A=120°, AB=AC=1 Find: The radius of circumscribed circle

Citrus2011 [14]

In Δ ABC, ∠A=120°, AB=AC=1

To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.

Then join OA, OB, and OC.

Taking any one OA,OB and OC as radius draw the circumcircle.

Now, from O Draw OM⊥AB and ON⊥AC.

As chord AB and AC are equal,So OM and ON will also be equal.

The reason being that equal chords are equidistant from the center.

AM=MB=1/2 and AN=NC=1/2 [ perpendicular from the center to the chord bisects the chord.]

In Δ OMA and ΔONA

OM=ON [proved above]

OA is common.

MA=NA=1/2 [proved above]

ΔOMA≅ ONA [SSS]

∴ ∠OAN =∠OAM=60° [ CPCT]

In Δ OAN

$\cos60=\frac {AN}{OA}$

$\frac{1}{2}=\frac{\frac{1}{2}}{OA}$

OA=1

∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.

4 0

1 year ago