Pleas help me it hard for me
1 answer:
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Answer:
Kindly check attached picture and explanation
Step-by-step explanation:
A.) The tree diagram is attached in the picture below.
B.) Number less than 3 :
Event A:
A = {1HH, 1HT, 1TT, 1TH, 2H, 2T}
C.)
Event B:
B = {1TT, 3TT, 5TT}
D.)
Event C:
C = {1HT, 1TH, 3HT,3TH,5HT, 5TH}
E.)
A' = Elements not in A = number ≥ 3
A' = {3HH,3HT, 3TT, 3TH, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}
Question f and G are not well formatted ;
h. Check attached picture
I) i. Are A and B mutually exclusive? Are A and C mutually exclusive? Are B and C mutually exclusive?
A and B are mutually exclusive if AnB = ∅
Check :
AnB = {1TT}
Hence, A and B are not mutually exclusive
A and C
AnC = {1HT, 1TH}
A and C are not mutually exclusive.
B and C
BnC = ∅
B and C are mutually exclusive
Here, “p” is a numerator and “q” is a denominator. The examples of rational numbers are 6/5, 10/7, and so on. The rational number is represented using the letter “Q”. Like real numbers, the arithmetic operations, such as addition, subtraction, multiplication, and division are applicable to the rational numbers.
Define
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s