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2 years ago

Type the value of the ? below.

Mathematics

2 answers:

larisa [96]2 years ago

7 0

Answer:

416

Step-by-step explanation:

i geuss.

blsea [12.9K]2 years ago

3 0

Answer:

416

Step-by-step explanation:

ummmmmmmm yeah

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What is the value of x in the equation 1,331x with the exponent of 3 -216 =0

dimulka [17.4K]

The answer is x=6/11.
Tiger Algebra is a really good tool for algebra if you need some help.
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3 years ago

Simplify the expression:<br><br> –3(9 − x) =

steposvetlana [31]

Answer:

Expand $-3(9-x): -27 + 3x$

Step-by-step explanation:

$-3(9-x)$

Apply the distributive law: $a(b-c)=ab-ac$

$a = -3, b=9, c=x$

$=-3*9-(-3)x$

Apply minus - plus rules:

$-(-a)=a$

$= -3*9+3x$

Multiply the numbers: $3 * 9 = 27$

$=-27+3x$

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Thank you, have a good one! =)

6 0

2 years ago

What are the AGI and taxable income?

andreev551 [17]

Answer:

1. Find the AGI and taxable income: Gross Income $30,856 Adjustments $750 1 Exemption $8200 Deduction $2,300

AGI: $31,200 and $20,601 $30106 --- ANSWER: 30,106 (30,856-750)

Taxable Income: $19,606 $29,586 and $18,505 $28,863 and $17,636 1 points--- ANSWER 19,606

2. QUESTION 5 Find the AGI and taxable income. Gross Income $67,890

Adjustments $0 3 Exemptions $24,600 Deduction $1469

AGI: $69,440 and $45,300 $68,990 and $42,831 $67,890 --- ANSWER:

67,890

Taxable Income: $41,821 $65,551 and $44,821 1 points --- ANSWER: 41,821 (67,890-24,600-1,469)

3. QUESTION 6 Find the AGI and taxable income. Gross income $19,723 Adjustments $255 1 Exemption $8200 Deduction $1430 $19,4

AGI: 19,468 (19,723-255)

Taxable Income: 9,838 (19,468-8,200-1,430)

Step-by-step explanation:

Still stuck? Get 1-on-1 help from an expert tutor now.

3 0

2 years ago

Read 2 more answers

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$