When dealing with little odds you must know one part and the whole part. For number 1 a fraction , 7 is a part of a 11 which is the whole. For 2 a percent, 6% part and 100% whole, here you had to know that for 6% to exist there had to be a 100% . For 3 a ratio, the first number is part (2) and the second is whole (7). Look at pic for all work.
Answer:
The answer here is the third box(C)
I hope this helps you!
Answer: 7
Step-by-step explanation:
14*1=14
14/2=7
Answer:
n > -2
Step-by-step explanation:
7(1 - 2n) < 33 - n
7 - 14n < 33 - n
Subtract 7
-14n < 26 - n
Add n
-13n < 26
Divide by -13
n > -2
Given:
p = 90% = 0.9, the probability that an adult has had chickenpox by age 50.
Therefore,
q = 1 - p = 0.1, the probability that an adult has not had chickenpox by age 50.
Part (a)
Because there are only two answers: "Yes" or "No" to whether an adult has had chickenpox by age 50, the use of the binomial distribution is justified.
Part (b):
Calculate the probability that exactly 97 out of 100 sampled adults have had chickenpox.
The probability is
P₁ = ₁₀₀C₉₇ (0.9)⁹⁷ (0.1)³ = 0.0059
Answer: 0.006 or 0.6%
Part (c)
Calculate the probability that exactly 3 adults have not had chickenpox.
Theis probability is equal to
P₂ = 1 - P₁ = 1 - 0.006 = 0.994
Answer: 0.994 or 99.4%
Part (d)
Calculate the probability that at least 1 out of 10 randomly selected adults have had chickenpox.
The probability is
P₃ = ₁₀C₀ (0.9)⁰ (0.1)¹⁰ + ₁₀C₁ (0.9)¹ (0.1)⁹ = 10⁻¹⁰ + 10⁻⁹ = 10⁻⁹ ≈ 0
Answer: 0
Part (e)
Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.
The probability is
P₄ = 1 - [₁₀C₀ (0.9)⁰(0.1)¹⁰ + ₁₀C₁ (0.9)¹(0.1)⁹ + ₁₀C₂ (0.9)²(0.1)⁸ + ₁₀C₃ (0.9)³(0.1)⁷]
= 1 - (10⁻¹⁰ + 9 x 10⁻⁹ + 3.645 x 10⁻⁷ + 8.748 x 10⁻⁶)
= 1
Answer: 1.0 or 100%