All categories

3 years ago

What are the coordinates of the fourth point that could be connected with (–8, 0), (1, 0), and (1, –5) to form a rectangle?

1 answer:

6 0

(-8, -5)

To form a rectangle, all the lines must be perpendicular and parallel. If you draw out the points on a coordinate plane then you can find the answer pretty easily.

Also, this one is easy because all the lines are vertical or horizontal, and when that is the case then you can just see which number in the set hasn't appeared twice (1 appeared twice as an x value, 0 appeared twice as a y value, but -8 has only appeared once as an x and -5 has only appeared once as a y value)

You might be interested in

F(x)=9x^2 g(x)= sqrt 12-x/3 (f o g)(-4)

Slav-nsk [51]

$f(x)=9x^2 \\ g(x)=\frac{\sqrt{12-x}}{3} \\$ So,first step is to write $(fog)(-4)) =f[g(-4)] \\$

Now we start from inner paranthesis ,we need to first find value of $g(-4) =\frac{\sqrt{12-(-4)}}{3}\\ =\frac{\sqrt{12+4}}{3}\\
=\frac{\sqrt{16}}{3}\\
=\frac{4}{3}\\
(fog)(-4)) =f[g(-4)] =f(\frac{4}{3}) =9(4/3)^{2} =9*(16/9) =144/9 =16$

8 0

3 years ago

A vacuum cleaner costs £85

Vitek1552 [10]

Answer: £76.50

Step-by-step explanation:

You find 10% of 85 = 8.5

then you do £85-£8.50 = 76.50

7 0

2 years ago

The work shows how to use long division to find?

KIM [24]

Answer:

remainder = $\frac{1}{x-2}$

Step-by-step explanation:

The remainder is the value of the subtraction of the last 2 lines in the procedure.

5x - 9

- (5x - 10)

That is

5x - 5x = 0 and - 9 - (- 10) = - 9 + 10 = 1

over the divisor x- 2 , gives

remainder = $\frac{1}{x-2}$

6 0

2 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

What 1,112,433 rounded to the nearest ten thousand

ValentinkaMS [17]

1,110,000.
You look at the ten thousands value and look at the number after it. If it is less than 5 then it is that number, just add zeros to the rest of the values after it.

5 0

3 years ago