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3 years ago

Complete the following statement. Use the integers that are closest to the number in the

Mathematics

1 answer:

aivan3 [116]3 years ago

8 0

Answer:

is this deleted or not siris this deleted or not sir

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3х – Зу = 15<br> Determine if the given ordered pair, (5,4) satisfies the given equation

galina1969 [7]

Answer:

it does not.

Step-by-step explanation:

3(5)-3(4)=15

15-12=15

3=15

6 0

3 years ago

Henry learned 21 new vocabulary words. That is 3 times as many

Novosadov [1.4K]

Answer:

Step-by-step explanation:

21 divided by 3 = 7

7 x 3 = 21

5 0

3 years ago

4.5kg of oranges and 3kg of apples cost a total of £3.87 2.5kg of oranges cost £1.40 Work out the cost of 1kg of apples.

PolarNik [594]

Answer:

=£0.516 is 1 kg of apples

5 0

3 years 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

3 years ago

George believes the Art Club students at his school have an unfair advantage in being assigned to the art class they request. He

Whitepunk [10]

Answer:

class assignment seems to discriminate against Art Club members

Step-by-step explanation:

The probability of getting the requested assignment given that the student was in Art Club was ...

P(got class | art club) = P(got class & art club)/P(art club)

P(got class | art club) = (100/500)/(165/500) ≈ 60.6%

P(got class | not club) = P(got class & not club)/P(not club)

P(got class | not club) = (265/500)/(335/500) ≈ 79.1%

It looks like George's theory may be incorrect, as a significantly greater percentage of non-club students got their requested class than did club students. The class assignment process and art club membership are not independent, suggesting an unfair disadvantage for club students.

4 0

3 years ago