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Anna [14]
3 years ago
9

HELO FASTTTTTT PLEASEEEEE DJEIFJEJCINWKDEMDKENDKWKDS

Mathematics
2 answers:
IrinaK [193]3 years ago
5 0

Answer:

I think it's all of them :)

MatroZZZ [7]3 years ago
5 0
All of them are real
You might be interested in
Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
3 years ago
HELP PLEASE
vredina [299]
The answer to you question is B for number 1 and for 2 it is C. Hope this helps
4 0
3 years ago
Read 2 more answers
Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:
tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

  • 91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

  • 79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

Q1 = \frac{N +1}{4}th

So, we have:

Q1 = \frac{13 +1}{4}th

Q1 = \frac{14}{4}th

Q1 = 3.5th

This is the average of the 3rd and the 4th element

Q1 = \frac{1}{2} \times (84 + 91)

Q1 = 87.5

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

Q3 =  3 \times \frac{N +1}{4}th

So, we have:

Q3 =  3 \times \frac{13 +1}{4}th

Q3 =  3 \times 3.5th

Q3 =  10.5th

This is the average of the 10th and the 11th element.

Q_3 =\frac12 \times (98 + 101)

Q_3 =99.5

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

IQR = Q_3 -Q_1

IQR = 99.5 - 87.5

IQR = 12

Also, we have:

IQR(1.5) = 12 \times 1.5

IQR(1.5) = 18

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

Lower = Q_1 - IQR(1.5)

Lower = 87.5- 18

Lower = 69.5

Upper = Q_3 + IQR(1.5)

Upper = 99.5 + 18

Upper = 117.5

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0
2 years ago
The length of a rectangle is twice it’s width. If the area of the rectangle is 72ft ^2, find it’s perimeter
alisha [4.7K]

Hi!

length = l = 2 * width = 2 * w

Area = l * w = l * 2 * l = 2 * l^2

2 * l^2 = 72

l^2 = 72/2 = 36

l = square root of 36

l = 6

l = 2 * w => w = l/2 = 6/2 = 3

Perimeter = 2 * l + 2 * w = 2 * 6 + 2 * 3 = 12 + 6 = 18 feet

Hope this helps!

7 0
3 years ago
How to factor quadratic with a leading coefficient?
Harrizon [31]
One method is called the ac method:-

For  example  factor

3x^2 + 2x - 21

ac ( first coefficient times the last) = 3*21 = -63

now we need to find 2 factors of -63 which add up to the coefficient of x  ( + 2):-

+9 and - 7 look good so we can write the original expression as:-

3x^2 + 9x - 7x - 21

Factor by grouping:-

= 3x(x +3) - 7(x + 3)     (  x + 3) is common so:-

= (3x - 7)(x + 3) Answer




7 0
3 years ago
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