Can someone help me on number one and 4? Thanks! :)

How do I determine whether 3x-4y+7=3y is a linear equation and how do I write it in standard form?? (I learned this already but

Pani-rosa [81]

Equation of the straight line can be given in more form. The most common forms are implicit (general or standard) form ax+by+c=0 and explicit form y=kx+i, where k is line coefficient and l is cut which line made on the y axis. If k>0 then the angle that takes straight line with the positive direction to the x axis is sharp and if k<0 then the angle that takes straight line with positive direction to the x axis is obtuse. In you case you only need to form one monomial with variable y in the given equation in the following way: 3x-4y+7=3y => add to both side (-3y) and you get 3x-4y-3y+7=3y-3y finally we get implicit or general 3x-7y+7=0. If is it necessary to transform from the implicit into the explicit form we will do this in the following way: 3x-7y+7=0 add to both side expression (-3x-7) => 3x-3x-7y+7-7=-3x-7 => divide both side with (-7) => y= (-3x-7)/ (-7) => finally we get y=3/7 x + 1 ( in our case coefficient of direction k=3/7 and the cut which line is made3 on the y axis l=1). Its display in the decartes coordinate system is given in one of the already given answers.

3 0

4 years ago

Can anyone help me with this? i don’t understand it, thank you.

Alina [70]

Answer:

24

Step-by-step explanation:

It is done by Middle term splitting.

Taking the general equation ax² + bx + c, by middle term splitting, "b" should be splitted such that the numbers either on adding or on subtracting must be equal to "b" and their products must be equal to "a x c".

For example, in the question it's already given that the equation 8x² + bx + c is middle term splitted into 8x² + px + qx + 3.

So, that means "p x q" must be equal to "8 x 3", i.e., 24.

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>

7 0

2 years ago

In a regression class, a student suggested the following: If extremely influential outlying cases are detected in a data set, si

mel-nik [20]

Answer:

Discarding the influential outlying cases when detected is also known as flagging outliers in a data set, and this is because outliers do not follow the rest of the dataset's pattern. if this outliers are not discarded they would have a negative effect on any model attached to the dataset

Step-by-step explanation:

In a regression class ; If extremely influential outlying cases are detected in a Data set, discarding this influential outlying cases is the right way to go about it

Discarding the influential outlying cases when detected is also known as flagging outliers in a data set, and this is because outliers do not follow the rest of the dataset's pattern. if this outliers are not discarded they would have a negative effect on any model attached to the dataset

6 0

3 years ago

In what line do planes A and QRV intersect?

matrenka [14]

In this question, we have to fin the line of intersection of the planes A and QRV.

We have to see the given figure and find the line on which the two planes A and QRV meets.

Lines SR,TS, WT and WQ are not on plane QRV.

The only line which passes through both planes is QR.

And that's the line of intersection of the two planes.

6 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago