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2 years ago

Any help will be really appreciated

Mathematics

2 answers:

RSB [31]2 years ago

5 0

Answer: B

Step-by-step explanation:

matrenka [14]2 years ago

3 0

The answer is B

5x2=10. 10+1=11
7x2=14. 14+1=15
9x2=18. 18+1=19

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The weights of boxes of candies produced in a factory are normally distributed with a mean weight of 16 oz and a standard deviat

JulijaS [17]

Let the required weight be X, then
z = (X - mean)/standard deviation
2 = (X - 16)/1
X - 16 = 2
X = 2 + 16
X = 18 oz.

3 0

2 years ago

Read 2 more answers

(6, - 3, 1) and (8,9,-11) find the angle between the vector

Svetllana [295]

Answer:

The angle between the two vectors is 84.813°.

Step-by-step explanation:

Statement is incomplete. Complete form is presented below:

<em>Let be (6,-3, 1) and (8, 9, -11) vector with same origin. Find the angle between the two vectors. </em>

Let $\vec u = \langle 6, -3, 1 \rangle$ and $\vec v = \langle 8,9,-11 \rangle$ , the angle between the two vectors is determined from definition of dot product:

$\theta = \cos^{-1} \left(\frac{\vec u \,\bullet \,\vec v}{\|\vec u\|\cdot \|\vec v\|} \right)$ (1)

Where:

$\vec u$ , $\vec v$ - Vectors.

$\|\vec u\|$ , $\|\vec v\|$ - Norms of each vector.

Note: The norm of a vector in rectangular form can be determined by either the Pythagorean Theorem or definition of Dot Product.

If we know that $\vec u = \langle 6,-3,1 \rangle$ and $\vec v = \langle 8, 9,-11 \rangle$ , then the angle between the two vectors is:

$\theta = \cos^{-1}\left[\frac{(6)\cdot (8) + (-3)\cdot (9) + (1)\cdot (-11)}{\sqrt{6^{2}+(-3)^{2}+1^{2}}\cdot \sqrt{8^{2}+9^{2}+(-11)^{2}}} \right]$

$\theta \approx 84.813^{\circ}$

The angle between the two vectors is 84.813°.

6 0

2 years ago

How many solutions does this equation have?<br> 9c − 10 = –10c − 10

Marina86 [1]

2 solutions because you have 2 different equations

6 0

2 years ago

(Please answer before 5:15 P.M tonight)

taurus [48]

The scale factor of dilation from ABC to $A^{1} B^{1} C^{1}$ is 0.333.

Step-by-step explanation:

Step 1:

First, we plot the coordinates of both triangles.

The coordinates of the triangle $A^{1} B^{1} C^{1}$ are as follows;

$A^{1}$ = (2, 3), $B^{1}$ = (2, 1), and $C^{1}$ = (3, 1).

The coordinates of triangle ABC are;

A = (4, 9), B = (4, 3), and C = (7, 3).

Step 2;

Next, we need to find the distances between at least two coordinates for both triangles.

The distances between vertical lines are found by calculating the difference in y values while for horizontal lines, the distances are found by calculating the difference between the x values.

The distance of $A^{1} B^{1} = 3-1=2$ , the distance of $B^{1} C^{1} = 3-2=1$ .