250 students complete a business college diploma. The results are normally distributed with a mean of 64% and a standard deviation of 8.6%. 15 students gain distinctions, 35 gain credits and 23 fail the course. Determine the cut-off percentages for each of the grades: a) Pass b) Credit c) Distinction

fail 0-x. mid values. =X/2

pass x-a. =(x+a)/2. pass no.=250-35-15-23=177

credit a-y. =(a+y)/2

distinction y-100. =(y+100)/2

mean=1/250*(x/2)×23+(x+a)/2*177+(y+a)/2×35+(y+100)/2×15=64

1/500[23x+177x+177a+35a+35y+15y+100*15)=64

200x+212a+50y+1500=32000

variance=[(X/2-64)^2*23+(x+a,/2-64)^2*277+(a+y/2-64)^2*35+(100+y/2-64)^2*15]/250=(8.6)^2

I've made two equations that has to be simplified.

probably a nother data has to be given since we have 3 variables to solve for(X,a,y)

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.