in an esp experiment subjects must predict whether a number randomly generated by a computer will be odd or even. What is the probability that a subject would guess more than 20 correct in a series of 36 trials?

it has 4

step-by-step explanation:

les

elee

fg

elle

fetish

ser

el

le

f

lent

clee hd

lg

ad

be

hu

te

see

te

ele

cele

her

t

errel

c

s​

sasaasasa

step-by-step explanation:

sasasdfubisdbjvkzc

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x