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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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Cubed root x cubed root x2​
Yuki888 [10]

Answer:

Final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

Hence final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

7 0
3 years ago
An envelope contains three cards: a black card that is black on both sides, a white card that is white on both sides, and a mixe
Over [174]

Answer:

There is a  2/3  probability that the other side is also black.

Step-by-step explanation:

Here let B1: Event of picking a card that has a black side

B2: Event of picking a card that has BOTH black side.

Now, by the CONDITIONAL PROBABILITY:

P(B_2/B_1 )  = \frac{P(B_1\cap B_2)}{P(B_1)}

Now, as EXACTLY ONE CARD has both sides BLACK in three cards.

⇒ P (B1 ∩ B2) = 1 /3

Also, Out if total 6 sides of cards, 3 are BLACK from one side.

⇒ P (B1 ) = 3 /6 = 1/2

Putting these values in the formula, we get:

P(B_2/B_1 )  = \frac{P(B_1\cap B_2)}{P(B_1)} = \frac{1}{3}  \times\frac{2}{1} = \frac{2}{3}

⇒ P (B2 / B1)  =  2/3

Hence, there is a  2/3  probability that the other side is also black.

 

5 0
3 years ago
Question 1 help!? Asap
Paha777 [63]

Hi!

If the perimeter of the triangle is 41, then x is:

5x + 2 + 6x - 3 + 3x = 41

5x + 6x + 3x = 41 + 3 - 2

14x = 42

x = 42/14

x = 3

The x is 3.

7 0
3 years ago
What is the solution to 4x-3y=10
Ainat [17]
Solving for y:
4x-3y=10. (subtract 4x on both sides)
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y=4/3 x+10/3
6 0
3 years ago
The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi
Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

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Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Now, z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0
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