Mathematics, 13.03.2020 02:56 DnsMonsteR7712

A test consists of 10 true/false questions. To pass the test a student must answer at least 6 questions correctly. If a student guesses on each question, what is the probability that the student will pass the test?

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Answer from: savannahsharp5149

37.70% probability that the student will pass the test

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student guesses it correctly, or he does not. The probability of a student guessing a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

10 true/false questions.

10 questions, so n = 10

True/false questions, 2 options, one of which is correct. So $p = \frac{1}{2} = 0.5$

If a student guesses on each question, what is the probability that the student will pass the test?

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051$

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

$P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439$

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.0010$

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2051 + 0.1172 + 0.0439 + 0.0098 + 0.0010 = 0.3770$

37.70% probability that the student will pass the test

Answer from: Quest

62926281916 inches of rain: )

Answer from: Quest

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

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