(a)/(2a)+(4)/(3a^(2)) ayuda por favor

Does anyone know what is the measure for these need help on this

sasho [114]

4. 66
5. 48
6. 60
7. 108
8. 60

4 0

3 years ago

Kelly, Jess, Sam and Evan are running for student council. Show all the ways that a president and vice president can be elected.

djyliett [7]

Kelly, Jess
Kelly, Sam
Kelly, Evan
Jess, Sam
Jess, Evan
Sam, Evan

4 0

3 years ago

jennifer made $74,900 last year she pay 6% state income tax, 15% federal income tax 6.2% for social security and 1.45% for medic

Gennadij [26K]

The idea here is that $74,900 is Jennifer's annual salary before the deductions (the taxes, social security and medicare). Since she pays 6% in state taxes, 15% in federal taxes, 6.2% in social security and 1.45% in medicare, the deductions total 6% + 15% + 6.2% + 1.45% = 28.65%. To find out what she keeps we do the total (100%) less the deductions (28.65%) which yields 71.35%. That is, she keeps 71.35% of the total amount given ($74,900).

Recall, that percent means out of ("per") 100 ("cent" --> think "century= 100 years). So, 71.35% means 71.35 out of 100. That is $\frac{71.35}{100}=.7135$ . To find 71.35% of 74,900 we multiply $(.7135)(74900)=53,441.15$ . That means that Jennifer keeps $53,441.15 of what she makes.

We are asked for her monthly net income. That is, how much money does she take home per month after the deductions. If we assume that she is paid the same amount each month, to find her monthly net income we divide 53,441.15 by 12.

Jennifer's net monthly income is $\frac{53441.15}{12}=$4,453.43$

Note that when we divide by 12 above we actually get 4,453.42916 but as this is money I rounded to $4,453.43. In reality this is a little bit too high as 4,453.43 times 12 = 53,441.16. That means we overpaid her by one penny. In real life she probably would get 11 paychecks at $4,453.43 and one paycheck at $4,453.42 to make up for the penny, but if this is a textbook problem I expect the rounding up would be acceptable.

5 0

3 years ago

The annual expenditure of the US federal government is approximately 4 44 trillion dollars. If a one dollar bill is 0 . 0 0 0 1

kaheart [24]

Answer:

<em>A stack of 4 trillion one dollar bills will be $4\times 10^8$ meters.</em>

Step-by-step explanation:

The annual expenditure of the US federal government is approximately 4 trillion dollars.

We know, 1 trillion dollars = 10¹² dollars.

So, 4 trillion dollars $=(4\times 10^1^2)$ dollars.

Each one dollar bill is 0.0001 meters thick.

So, the total height of the stack of 4 trillion one dollar bills will be.....

$[0.0001\times(4\times 10^1^2)]meters\\ \\ =[10^-^4\times(4\times 10^1^2)]meters\\ \\ =(4\times 10^8) meters$

6 0

3 years ago

A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),

lana [24]

Answer:

(9) $\frac{1}{12}$ (10) $\frac{1}{12}$ (11) $\frac{5}{12}$ (12) $\frac{1}{4}$ (13) $\frac{1}{6}$ 14) $\frac{5}{36}$ (15) $\frac{1}{12}$ (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are: (3, 6), (4, 5), (5, 4)

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are: (1, 3),(2, 2),(3, 1),

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

$P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}$

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

$P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}$

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

$P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}$

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

$P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}$

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

$P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}$

16.Probability of getting two numbers with a sum of 14.

$P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0$

4 0

4 years ago