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scZoUnD [109]
3 years ago
7

Mike has a net spendable income of $1,400. He decides to set up a budget before looking for an apartment or a car. He sets up hi

s budget and finds that he has lots of money left over. He puts the extra money into entertainment.
Housing $420
Food $168
Transportation $210
Insurances $42
Debts $70
Entertainment $224
Clothing $70
Savings $70
Medical $56
Miscellaneous $70
Why might this budget be a problem?


A.) Mike will spend more money than he is earning.

B.) Mike does not have enough money budgeted for food or medical expenses.

C.) Mike is using the minimum recommended percentages for food, housing, and transportation. The costs are likely to exceed the expenses in each category.

D.) Mike has not budgeted all of his money. He will find that he has extra money at the end of the month, which will end up being wasted.
Mathematics
2 answers:
Shtirlitz [24]3 years ago
7 0

C.) Mike is using the minimum recommended percentages for food, housing, and transportation. The costs are likely to exceed the expenses in each category.


Triss [41]3 years ago
5 0

Answer:

Option C is the answer.

Step-by-step explanation:

Mike has a net spendable income of $1,400.

His total budget is :

Housing =$420

Food=$168

Transportation=$210

Insurances=$42

Debts=$70

Entertainment=$224

Clothing=$70

Savings=$70

Medical=$56

Miscellaneous=$70

420+168+210+42+70+224+70+70+56+70=1400

So, out of all the given options, the correct one seems to be :

Mike is using the minimum recommended percentages for food, housing, and transportation. The costs are likely to exceed the expenses in each category.

Therefore, we can say that Mike should budget properly as he is spending more on entertainment, rather than more important categories.

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cos(x) = -\frac{1}{2}

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Since only 120° is an option, the answer is B.
7 0
3 years ago
PLEASE HELP
Nookie1986 [14]

For the square with side length n, the diagonal measures:

d = \sqrt{2} *n

<h3>How to get the length of the diagonal?</h3>

The sidelength of the square is n, and we want to get the length of the diagonal d.

Notice that the diagonal is the hypotenuse of a right triangle whose catheti measure n.

Then we can use the Pythagorean theorem, which says that the square of the hypotenuse is equal to the sum of the squares of the cathetus;

d^2 = n^2 + n^2\\\\d^2 = 2n^2\\\\d = \sqrt{2n^2} \\\\d = \sqrt{2}*n

That is the length of the diagonal.

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6 0
2 years ago
What is the equation of the line passing through (-5, - 3)
Leona [35]

Answer:

y=7/5x+4

Step-by-step explanation:

the other ones either are to low on the graph to hit -5-3 or they are parallel instead of perpendicular to the given equation

6 0
3 years ago
Investigate the difference between compounding annually and simple interest
Gennadij [26K]

Step-by-step explanation:

Simple interest formula

A = P (1 + rt)

Compound interest formula

A = P(1 + \frac{r}{n})^{nt}

a.

A = 5000 (1 + 0.025*1)\\A=5000(1.025)\\A=5125

Simple interest is $125

b

. A = 5000 (1 + \frac{0.025}{1})^{1*1}      \\A=5000(1.025)\\A= 5125

Compound interest is $125

c. the result for both a and b are the same

d.

A = 5000 (1 + 0.025*3) \\A=5000(1.075) \\A=5375

the simple interest is $375

e

. A = 5000 (1 + \frac{0.025}{1})^{1*3}] \\A=5000(1.025)^3 \\A=5000(1.077)\\A= 5385

the compound interest is $385

f. the result compared, compound interest is $10 more than simple interest

g.

A = 5000 (1 + 0.02*6) \\A=5000(1.12) \\A=5600

the simple interest is $600

h.

A = 5000 (1 + \frac{0.02}{1})^{1*6}] \\A=5000(1.12)^6 \\A=5000(1.9738) \\A= 9869

the compound interest is $4869

i. the result from g and h, h is over 8 times bigger than g.

j. interest compound annually is not the same as simple interest, only for the case of a and b seeing that it is for 1 year. but for 2years and above there is difference as seen in c to h

6 0
3 years ago
What are the greatest numbers : 4, -8, 4, 8, 7, -3, -9, 2, 6, 1, -5, 5, -2, 0.
myrzilka [38]

Answer:

Step-by-step explanation:

I would say 8 is the greatest number

the order from least to greatest

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7 0
3 years ago
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