Given a normal distribution with image for (a)the normal -curve area to the right of x=17. (b)the normal-curve area to the left of x=22. (c)the normal-cu and image for (a)the normal -curve area to the right of x=17. (b)the normal-curve area to the left of x=22. (c)the normal-cu , find

(a)the normal -curve area to the right of x=17.
(b)the normal-curve area to the left of x=22.
(c)the normal-curve area between x=32 and x=41
(d)the value of x that has 80% of the normal-curve area to theleft.
(e)the two values of x that contain the middle 75% of thenormal-curve area

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

i think it is c, not positive though

step-by-step explanation:

oof

step-by-step explanation:

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.