Find the indicated probability. round to three decimal places. a test consists of 10

a. true

b. 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?

To answer this problem, we use the binomial distribution formula for probability:

P (x) = [n! / (n-x)! x!] p^x q^(n-x)

Where,

n = the total number of test questions = 10

x = the total number of test questions to pass =   >6

p = probability of success = 0.5

q = probability of failure = 0.5

Given the formula, let us calculate for the probabilities that the student will get at least 6 correct questions by guessing.

P (6) = [10! / (4)! 6!] (0.5)^6 0.5^(4) = 0.205078

P (7) = [10! / (3)! 7!] (0.5)^7 0.5^(3) = 0.117188

P (8) = [10! / (2)! 8!] (0.5)^8 0.5^(2) = 0.043945

P (9) = [10! / (1)! 9!] (0.5)^9 0.5^(1) = 0.009766

P (10) = [10! / (0)! 10!] (0.5)^10 0.5^(0) = 0.000977

Total Probability = 0.376953 = 0.38 = 38%

There is a 38% chance the student will pass.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.