Let’s go through each question one by one:
### Question 17:
There are 15 white and 25 black identical balls in a box. To find the probability that a ball selected at random from the box is white, we follow these steps:
1. Count the total number of balls:
The total number of balls is [tex]\(15 + 25 = 40\)[/tex].
2. Determine the number of favorable outcomes:
The favorable outcomes are the selections of a white ball. There are 15 white balls.
3. Calculate the probability of selecting a white ball:
The probability of an event is given by the formula:
[tex]\[
P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\][/tex]
Here, the probability that the ball is white is:
[tex]\[
P(\text{White}) = \frac{15}{40} = \frac{3}{8}
\][/tex]
Therefore, the answer is:
[tex]\[
\boxed{\frac{3}{8}}
\][/tex]
### Question 18:
Given the vectors [tex]\( m = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \)[/tex] and [tex]\( n = \begin{pmatrix} -4 \\ 2 \end{pmatrix} \)[/tex], we need to find [tex]\( 2m + n \)[/tex].
1. Calculate [tex]\( 2m \)[/tex]:
[tex]\[
2m = 2 \begin{pmatrix} 5 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \times 5 \\ 2 \times -1 \end{pmatrix} = \begin{pmatrix} 10 \\ -2 \end{pmatrix}
\][/tex]
2. Add vector [tex]\( n \)[/tex] to [tex]\( 2m \)[/tex]:
[tex]\[
2m + n = \begin{pmatrix} 10 \\ -2 \end{pmatrix} + \begin{pmatrix} -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 10 + (-4) \\ -2 + 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}
\][/tex]
Therefore, the answer is:
[tex]\[
\boxed{\begin{pmatrix} 6 \\ 0 \end{pmatrix}}
\][/tex]
### Question 19:
If the gradient (slope) of a straight line is zero, then the line is:
- The gradient of a line indicates its steepness. A gradient of zero means no steepness.
- This implies that the line is perfectly flat and does not rise or fall as you move along it.
- Such a line is characteristic of a horizontal line, where [tex]\( y \)[/tex] remains constant for any [tex]\( x \)[/tex].
Therefore, the answer is:
[tex]\[
\boxed{\text{B. is horizontal}}
\][/tex]
### Question 20:
To find the image of the point [tex]\( (-3, 5) \)[/tex] when it is rotated through [tex]\( 360^\circ \)[/tex] about the origin, we analyze the transformation:
- A rotation of [tex]\( 360^\circ \)[/tex] brings the point back to its original position because one complete rotation returns all points in a plane to their starting positions relative to the origin.
Therefore, the image of the point [tex]\( (-3, 5) \)[/tex] remains [tex]\( (-3, 5) \)[/tex].
The answer is:
[tex]\[
\boxed{\text{B. } (-3, 5)}
\][/tex]
