The scores of a certain population on the Wechsler Intelligence Scale for Children (WISC) are thought to be Normally distributed with mean μ and standard deviation σ = 10. A simple random sample of 25 children from this population is taken and each is given the WISC. The mean of the 25 scores is = 104.32. Suppose a histogram of the 25 WISC scores is the following:
Based on this histogram, we would conclude that a 95% confidence interval of 104.32 ± 3.92 is:.
a) actually a 90% confidence interval.
b) actually a 99% confidence interval.
c) not very reliable.
d) very reliable.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

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.

since no one answered yet i will give it a shot, not completely sure tho.

i think it should be y=cos(x+2pi), (third answer, answer c)

good luck.