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3 years ago

8

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

3 0

3
—
-2
because the answer is 6 over -4 and it simplified is that answer.

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the doctor recommended that Matt increase the number of calories you consume from 2000 to 2800 by what percent did the number of

nikdorinn [45]

Answer:

40% number of calories have increased

Step-by-step explanation:

Old Value ( Calories consume before) = 2000

New Value (increase in Calories consumed) = 2800

We need to find by what percent did the number of calories increase

To find percent increase, the formula used is:

$Percent\: increase=\frac{New\:Value-Old\:Value}{Old\:Value}\times 100%$

Putting values in formula and finding percent increase

$Percent\: increase=\frac{New\:Value-Old\:Value}{Old\:Value}\times 100\%\:\\Percent\: increase=\frac{2800-2000}{2000}\times 100\%\: \\Percent\: increase=\frac{800}{2000}\times 100\%\: \\Percent\: increase=0.4\times 100\%\: \\Percent\: increase=40\%$

So, 40% number of calories have increased

8 0

3 years ago

Write 16^8 as a power with a base of 4

ruslelena [56]

(4^2)^8 would be the answer 4^2=16 and 16^8 is the same thing

4 0

3 years ago

Read 2 more answers

(b-2)x= 8

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

(2-2)x = 8

(0)x = 8

x = 8/0

no solution

6 0

3 years ago

9. Draw the number line graph of 0.4.

Evgesh-ka [11]

The number line graph of 0.4 is here.

4 0

3 years ago

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,

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

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

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

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

6 0

3 years ago