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qaws [65]
2 years ago
8

A quiz consists of 20 multiple-choice questions, each with 5 possible answers. For someone who makes random guesses for all of t

he answers, find the probability of passing if the minimum passing grade is 50 %. Enter your answer in decimal form accurate to six decimal places.
Mathematics
1 answer:
Pavlova-9 [17]2 years ago
7 0

Answer:

0.00259

Step-by-step explanation:

Probability, p of making the correct guess ; 1 / 5 = 0.2

Passing grade = 50% ; 50% * 20 = 0.5 * 20 = 10

q = probability of not passing ;

q = 1 - p = 1 - 0.2 = 0.8

Using the binomial distribution formula :

P(x >=x) = nCx * p^x * (1 - p)^(n - x)

P(x >= 10) = p(x = 10) + p(x = 11)+... + p(x = 20)

P(x >= 10) = 10C20 * 0.2^10 * 0.8^10

P(x >=) 0.00259

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Murrr4er [49]

You'll have to c<span>ompass tip on A and draw a small ark with pencil approximately in the middle above AB line, now compass tip to point B and cross the ark you made previously.
Do the same on the opposite side without making any change to the compass 
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8 0
2 years ago
Find the midpoint m of the line segment joining the points p (-5,-3) = q and = (-1,-1)
babymother [125]
Hello, hope this makes some sense, the midpoint formula is what’s given in the picture and the rest is just substitution

5 0
1 year ago
The radius of a nitrogen atom is 5.6 × 10^-11 meters, and the radius of a beryllium atom is 1.12 × 10^-10 meters. Which atom has
Airida [17]
The answer is Beryllium atom has a larger radius by 2 times.

Nitrogen atom: 5.6 × 10⁻¹¹ = 5.6 × 10⁻¹⁻¹⁰ = 5.6 × 10⁻¹ × 10⁻¹⁰ = 0.56 × 10⁻¹⁰
Beryllium atom: 1.12 × 10⁻¹⁰

Since 1.12 is bigger than 0.56, then the radius of beryllium atom is larger of the radius of nitrogen atom. Let's see by how many times:
1.12 × 10⁻¹⁰ : 0.56 × 10⁻¹⁰ = 1.12 : 0.56 = 2

4 0
3 years ago
Expanding Logarithmic Expressions In Exercise, use the properties of logarithms to rewrite the expression as a sum, difference,
Leya [2.2K]

Answer:  \dfrac{1}{3}[\ln (x+1)+\ln(x-1)]

Step-by-step explanation:

Properties of logarithm :

  1. \log (ab)= \log a+\log b
  2. \log(\dfrac{a}{b})=\log a-\log b
  3. \log a^n= n\log a

The given expression in terms of Natural log : \ln (x^2 - 1)^{\frac{1}{3}}

This will become \dfrac{1}{3}\ln (x^2 - 1)      [ By using Property (3)]

=\dfrac{1}{3}\ln (x^2-1^2)

=\dfrac{1}{3}\ln ((x+1)(x-1))   [\because a^2-b^2=(a+b)(a-b)]

=\dfrac{1}{3}[\ln (x+1)+\ln(x-1)]   [ By using Property (1)]

Hence, the simplified expression becomes \dfrac{1}{3}[\ln (x+1)+\ln(x-1)] .

5 0
3 years ago
In a class of 20 students:
Artyom0805 [142]

Whats the question? I dont understand whats happenign here.

5 0
2 years ago
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