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3 years ago

Write the sum of three odd consecutive integers if the last one is m-2

Mathematics

1 answer:

larisa86 [58]3 years ago

8 0

Answer:

the sum is 3m-12

Step-by-step explanation:

The nunbers are:

$m-2\\m-4\\m-6\\the~sum~is:\\sum=(m-2)+(m-4)+(m-6)\\sum=m-2+m-4+m-6=m+m+m-2-4-6\\sum=3m-12$

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How can a rational number and irrational number when multiplied together make a irrational number ?

pochemuha

Answer:

"The product of a rational number and an irrational number is SOMETIMES irrational." If you multiply any irrational number by the rational number zero, the result will be zero, which is rational. Any other situation, however, of a rational times an irrational will be irrational

A better statement would be:

"The product of a non-zero rational number and an irrational number is irrational

6 0

3 years ago

A rectangle has a base of 5 cm and a height of (x + 2)cm. Its area is 20 cm ^ 2. What is the value of x?

Len [333]

So area is base x height:
5(x+2)=20
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3 0

4 years ago

a recipe uses 1 1/4 cups of milk to make 10 servings. if the same amount of milk is used for each serving how mant servings can

laiz [17]

Answer:

I believe its 36

Step-by-step explanation:

so what i id is a took apart the answer, if 1 1/4 if 10 cups the 1/4 1/4 1/4 1/4 1/4 would also be 10 cups, so now i know that 1/4 = 2 servings

eso i took 3. 1/4 and that gave me 6 and now add the previous which is 10 so its 16 and i have 2 cups left

from here i knew 1 cup and 1/4 is 10 so subtract 1/4 which is -2 servings and you have 8 and 8+8 is 16 so

then i added

16+16= 36

which leads to my final answer of 36

Im so sorry if im wronge but if its right please give brainlest

7 0

3 years ago

How do you determine if a given set of measurements works for the side lengths of a triangle? (It should be a number)

zaharov [31]

Add them all up and if they add up to 180 then your side lengths are good

5 0

4 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