9514 1404 393
Answer:
D'(2, -1)
Step-by-step explanation:
Dilation about the origin by factor k multiplies each coordinate by that factor:
(x, y) ⇒ (kx, ky)
For k = 1/3, this is ...
(x, y) ⇒ (x/3, y/3)
D(6, -3) ⇒ D'(6/3, -3/3) = D'(2, -1)
Answer:
25+10y
Step-by-step explanation:
The area of a rectangle is
A = l* w where l is the length and w is the width
A = 5(5+2y)
Distribute
A = 25+10y
Answer:
so it looks like you want to find a 98% confidence interval and using a T value. So we have our X bar plus or minus a T. Star value times the standard deviation divided by the square root of van. And you had a sample of size 17 which yielded a mean of 32 that it lowered it by 32 the standard deviation was 15. And so we need to find the T. Star value. Now, subtle hint if you don't have a table on you, which I don't and I could always look it up online, but if I want the 98% confidence interval, that means I want the area down at this table to be 980.1 and I want my degrees of freedom to be 16. So I'm actually going to use my inverse, go to second and distribution and use my inverse T button on my T. I 84 and I put the area as 0.1 and my degrees of freedom as 16. And and then just enter in and I find out what that T. Star value is. This value is negative 2.583 and this value is 2.583 So now I have my T. Star value. You can also look it up in your in a table so we know that we'll have 32 plus or minus that T star value 2.583 times the standard deviation which is 15 over the square root of an Okay. And so I'm going to type in that 30 to minus 2.583 times the 15 divided by the square root of 17. And try to type it in right? And that gives me, my lower limit is 22.603 And my upper limit now I can go back and just change that subtraction sign to addition sign. And that gives me up to 41.397 So there's the confidence interval, the 98% confidence interval. And yes, you would. Uh that confidence interval, the mean would be unknown. Mhm. The standard deviation of the population is unknown because of the small sample size. We would like to have an approximate normal distribution, which means we would like to see a graph of those 17 numbers to make sure that they're approximately normal. And definitely we would like to have that be a simple random sample and it's not going to be most likely a simple random sample, but hopefully it's at least representative of what a simple random sample would be. Mhm.
All I know is look it up because you are in college and I am in elementary school almost to middle.
