Answer:
The sweat chloride reference value is less than 30 mmol/L. A value of more than 60 mmol/L of chloride in the sweat is consistent with a diagnosis of cystic fibrosis. The values of 30-60 mEq/L may represent heterozygous carriers, these carriers cannot be accurately identified with a sweat chloride test.
Given that,
Point 1 = (-6,9)
Point 2 = (2,9)
To find,
The slope of a line that passes through the given pair.
Solution,
We have,
x₁ = -6, x₂ = 2, y₁ = 9, y₂ = 9.
The formula for the slope of a line is given by :
Hence, the slope of the line that passes through the given pair is 0.
Step-by-step explanation:
#Hopeitshelp
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%
