The event "Atleast once" is the complement of event "None".
So, the probability that Marvin teleports atleast once per day will the compliment of probability that he does not teleports during the day. Therefore, first we need to find the probability that Marvin does not teleports during the day.
At Morning, the probability that Marvin does not teleport = 2/3
Likewise, the probability tha Marvin does not teleport during evening is also 2/3.
Since the two events are independent i.e. his choice during morning is not affecting his choice during the evening, the probability that he does not teleports during the day will be the product of both individual probabilities.
So, the probability that Marvin does not teleport during the day = 
Probability that Marvin teleports atleast once during the day = 1 - Probability that Marvin does not teleport during the day.
Probability that Marvin teleports atleast once during the day = 
Answer:
Step-by-step explanation:
Given that,
The dimensions of a rectangular block = 4 cm x 2 cm x 1.5 cm
Mass, m = 93.6 g
We need to find the density of steel. We know that the density of an object is equal to the mass divided by volume. So,
So, the density of the steel is 
.
Answer:
Alex can drive 480miles in one day period.
Step-by-step explanation:
$162- $42 =$120
$120/0.25¢=480
480miles
Answer:
13.4%
Step-by-step explanation:
Use binomial probability:
P = nCr p^r q^(n-r)
where n is the number of trials,
r is the number of successes,
p is the probability of success,
and q is the probability of failure (1-p).
Here, n = 16, r = 2, p = 0.25, and q = 0.75.
P = ₁₆C₂ (0.25)² (0.75)¹⁶⁻²
P = 120 (0.25)² (0.75)¹⁴
P = 0.134
There is a 13.4% probability that exactly 2 students will withdraw.
Answer:
8 - 2π square units.
π/16 - 1/8 square units.
6π - 9√3 square units.
Step-by-step explanation:
The area of the square = 2√2 * 2√2
= 2*2*2
= 8.
The area of the circle = πr^2
= π * [ ( 2√2)/ 2) ]^2
= π (√2)^2
= 2π.
Second Question:
The area of the circle = π(1/2)^2 = π/4.
Finding the area of the square:
1^2 = 2x^2
x^2 = 1/2
So the area of the square = 1/2
So the area of the red part = 1/4 ( π/4 - 1/2).
= π/16 - 1/8.
Third question
Area of the circle = 6^2 * π = 36π.
Now 60 degrees is 1/6 of 360 degrees so the are of the sector is 6π.
The area of the segment = 6π - 0.5 * 6^2 sin 60
= 6π - 18√3/2
= 6π - 9√3 square units.
