Answer:
a) Calculate the probability that at least one of them suffers from arachnophobia.
x = number of students suffering from arachnophobia
= P(x ≥ 1)
= 1 - P(x = 0)
= 1 - [0.05⁰ x (1 - 0.05)¹¹⁻⁰
]
= 1 - (0.95)¹¹
= 0.4311999 = 0.4312
b) Calculate the probability that exactly 2 of them suffer from arachnophobia? 0.08666
= P(x = 2)
= (¹¹₂) x (0.05)² x (0.95)⁹
where ¹¹₂ = 11! / (2!9!) = (11 x 10) / (2 x 1) = 55
= 55 x 0.0025 x 0.630249409 = 0.086659293 = 0.0867
c) Calculate the probability that at most 1 of them suffers from arachnophobia?
P(x ≤ 1)
= P(x = 0) + P(x = 1)
= [(¹¹₀) x 0.05⁰ x 0.95¹¹] + [(¹¹₁) x 0.05¹ x 0.95¹⁰]
= (1 x 1 x 0.5688) + (11 x 0.05 x 0.598736939) = 0.5688 + 0.3293 = 0.8981
You can divide the two values A = diameter of bacterium B = diameter of virus A/B = (10^(-6))/(10^(-7)) A/B = 10^(-6-(-7)) A/B = 10^(-6+7) A/B = 10^(1) A/B = 10 Since the ratio of the two diameters is 10, this means that the diameter of the bacterium is 10 times greater than that of the virus.
Answer: the system of equations are
3x + 2y = 170
4x + 6y = 360
Step-by-step explanation:
Let x represent the price of a child's ticket in dollars.
Let y represent the price of an adult's ticket in dollars.
The Brown family paid 170 for 3 children and 2 adults. This would be expressed as
3x + 2y = 170
The Peckham family paid 360 for 4 children and 6 adults. This would be expressed as
4x + 6y = 360
