Given:
Last season Steven missed 24% of the free throws in his basketball games.
Free throws in the first two games of this season = 25
His missed free throw percentage holds from last season.
To find:
The number of throws Steven miss during these two games.
Solution:
His missed free throw percentage for this season = His missed free throw percentage of last season = 24%
Free throws in the first two games of this season = 25
So, the number of throws Steven miss during these two games is



Therefore, Steven miss 6 free throws during these two games.
(D)
Step-by-step explanation:
sin x = opp/hyp
= 40/41
Step-by-step explanation:
it's B they have the same y intercept I think
Answer:
The side length of the each areas are (a) 4 cm (b) 12 cm (c) 6 cm (d) 7 cm (e) 8 cm (f) 2 cm .
Step-by-step explanation:
Given:
The areas of the squares:
(a) 16 cm² (b) 144 cm² (c) 36 cm² (d) 49 cm² (e) 64 cm² (f) 4 cm².
Now, finding the side length of each squares.
So, putting the formula of area of square for getting the side length(a):
(a) Area = a²

Using square root both the sides we get:

Side length = 4 cm.
(b) Area = a²

Using square root both the sides we get:

Side length = 12 cm.
(c) Area = a²

Using square root both the sides we get:

Side length = 6 cm.
(d) Area = a²

Using square root both the sides we get:

Side length = 7 cm.
(e) Area = a²

Using square root both the sides we get:

Side length = 8 cm.
(f) Area = a²

Using square root both the sides we get:

Side length = 2 cm.
Therefore, the side length of the each areas are (a) 4 cm (b) 12 cm (c) 6 cm (d) 7 cm (e) 8 cm (f) 2 cm.
Answer:
D. 0.002
Step-by-step explanation:
Given;
total number of sample, N = 500 elements
50 elements are to be drawn from this sample.
The probability of the first selection, out of the 50 elements to be drawn will be = 1 / total number of sample
The probability of the first selection = 1 / 500
The probability of the first selection = 0.002
Therefore, on the first selection, the probability of an element being selected is 0.002
The correct option is "D. 0.002"