What are the solutions to the equation?

Veseljchak [2.6K]

I don't know. This seems tricky. But I do know that the probability would be 1 divided by the probability. So A and D is Incorrect. Theres never really a 50-50 chance with Locker doors so I'm gonna go with C.

I'm not guaranteed this is the answer. But I do Hope it helps. =)

7 0

2 years ago

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Give me one example of an item where you would need to find the area of a square. Also, Give me one example of an item where you

monitta

An example of an item where you would need to find the area of a square is a square table.

An example of an item where you would need to find the area of a rectangle is a phone that has a rectangular shape.

<h3>How to calculate the area?</h3>

It's important to note that the area of a square is the multiplication of its sides by itself. For example, if the side is 4cm, the area will be:

= 4²

= 4 × 4.

= 16cm²

The area of a rectangle will be:

= Length × Width

Assuming length and width are 5cm and 2cm. This will be:

= 5 × 2

= 10cm²

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6 0

6 months ago

The coaches at Xavier Elementary School bought cases of sports drinks for a field day. They bought 76 cases of drinks. Each case

solong [7]

Answer:

608

Step-by-step explanation:

There are a total of 1,824 drinks since there were 76 cases at 24 drinks a piece. This creates a total number of 24*76 = 1824.

If each student receives 3 then 1824/3 = 608. There were 608 students.

7 0

2 years ago

Looking at the repayment time of peer to peer users how would you assume it compares to that every payment via cash or check? wh

Rina8888 [55]

Peer-to-peer is good for fast payment but is lacking in the transfer of funds security as there is mostly made b/w people with one unknown history or background, most people meeting for the first time, Compared to cash or check which is handled physically. This is further explained below.

<h3>Looking at the repayment time of peer-to-peer users how would you assume it compares to that every payment via cash or check?</h3>

Generally, P2P, which stands for peer-to-peer, is a method of payment that enables users to make purchases without needing to have access to their financial institution's account information. The transfer is completed quickly and, in most cases, at no cost. There is a generational divide in terms of adoption, but the vast majority of Americans today make use of mobile payment applications.

In conclusion, Peer-to-peer transactions are typically conducted between people who do not know each other's histories or backgrounds, and the majority of transactions involve people who are meeting for the first time. This makes peer-to-peer transactions less secure than cash or checks, which are physically handled. Peer-to-peer transactions are useful for making payments quickly.

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7 0

1 year ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

2 years ago