Answer:

Step-by-step explanation:
The equation of the line through the point
&
can be represented by:

Making m the subject;

∴
we need to carry out the equation of the line through (0,1) and (1,2)
i.e
y - 1 = m(x - 0)
y - 1 = mx
where;

m = 1
Thus;
y - 1 = (1)x
y - 1 = x ---- (1)
The equation of the line through (1,2) & (4,1) is:
y -2 = m (x - 1)
where;


∴

-3(y-2) = x - 1
-3y + 6 = x - 1
x = -3y + 7
Thus: for equation of two lines
x = y - 1
x = -3y + 7
i.e.
y - 1 = -3y + 7
y + 3y = 1 + 7
4y = 8
y = 2
Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7
∴



![\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cint%5E2_1%20%20%5Cbigg%20%28%20%5Bxy%5E2%5D%5E%7B-3y%2B7%7D_%7By-1%7D%20%5Cbigg%20%29%20%5C%20dy)
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![\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cint%5E2_1%20%20%5Cbigg%20%28%5By%5E2%28-4y%2B8%29%5D%20%5Cbigg%20%29%20%5C%20dy)

![\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%5Cdfrac%7B%20-4y%5E4%7D%7B4%7D%2B%5Cdfrac%7B8y%5E3%7D%7B3%7D%20%5Cbigg%20%5D%5E2_1)
![\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-y%5E4%2B%5Cdfrac%7B8y%5E3%7D%7B3%7D%20%5Cbigg%20%5D%5E2_1)
![\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-2%5E4%2B%5Cdfrac%7B8%282%29%5E3%7D%7B3%7D%20%2B%201%5E4-%20%5Cdfrac%7B8%5Ctimes%20%281%29%5E3%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-16%2B%5Cdfrac%7B64%7D%7B3%7D%20%2B%201-%20%5Cdfrac%7B8%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-15%2B%20%5Cdfrac%7B64-8%7D%7B3%7D%5Cbigg%5D)
![\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20-15%2B%20%5Cdfrac%7B56%7D%7B3%7D%5Cbigg%5D)
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![\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]](https://tex.z-dn.net/?f=%5Ciint_D%208y%5E2%20%5C%20dA%20%3D8%20%5Cbigg%20%5B%20%20%5Cdfrac%7B11%7D%7B3%7D%5Cbigg%5D)

Answer: a. 1.981 < μ < 2.18
b. Yes.
Step-by-step explanation:
A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.
First, we calculate mean of the sample:


2.08
Now, we estimate standard deviation:


s = 0.1564
For t-score, we need to determine degree of freedom and
:
df = 12 - 1
df = 11
= 1 - 0.95
α = 0.05
0.025
Then, t-score is
= 2.201
The interval will be
± 
2.08 ± 
2.08 ± 0.099
The 95% two-sided CI on the mean is 1.981 < μ < 2.18.
B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.
well, we know that 213.90 includes the 15% gratuity, so if the meal price was say "x", namely that is the 100%, and we add 15% to that, then 100% + 15% = 115%, so 213.90 is really the 115%.
now, if 213.90 is the 115.%, what is "x"? the 100%.

so since we know the meal without the tip is 186 bucks, the tip was then 213.90 - 186 = 27.90.
but they got extra potato wedges with butter and the waitress was wearing her best hat, they added $20 more, so the tip ended up as 27.90 + 20 = 47.90.
now, if the cost of the meal was $186 and that is the 100%, what is 47.90 off of it in percentage?

Answer:
They both have H hoel
Step-by-step explanation:
\