The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
<h3>How to determine the number of ways</h3>
Given the word:
ESTABROK
Then n = 8
p = 6
The formula for permutation without restrictions
P = n! ( n - p + 1)!
P = 8! ( 8 - 6 + 1) !
P = 8! (8 - 7)!
P = 8! (1)!
P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1
P = 40, 320 ways
Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
Learn more about permutation here:
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The solutions appear to be
{π/2, 2π/3, 4π/3}.
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
24 + 0.44x = 19 + 1.69x
Subtract 19 from both sides
5 + 0.44x = 1.69x
Subtract 0.44x from both sides
5 = 1.25x
Divide by 1.25 on both sides
4 = x
Answer:
0.49
Step-by-step explanation:
P(warning or ticket) = P(warning) + P(ticket) − P(warning and ticket)
0.52 = 0.03 + P(ticket) − 0
P(ticket) = 0.49