In a population of 120 voters in random town, suppose 40% will vote for phred for mayor. if you randomly sample 25 voters (without replacement) before the election, what’s the chance that the number in the sample who will vote for phred will be between 9 and 11 (i. e., within 4% of the true value)? do this using the exact (hypergeometric) and approximate (normal, with continuity correction) distributions.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

sorry man i accidentally pressed yes my bad

step-by-step explanation:

1. perpendicular

2. parallel

3. perpendicular