Answer:
The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>
Step-by-step explanation:
The given data are;
Bakery A
<em> </em>= 1,880 cal
s₁ = 148 cal
n₁ = 10
Bakery B
<em> </em>= 1,711 cal
s₂ = 192 cal
n₂ = 10
![\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}](https://tex.z-dn.net/?f=%5Cleft%20%28%5Cbar%7Bx%7D_1-%5Cbar%7Bx%7D_%7B2%7D%20%20%5Cright%20%29%20-%20t_%7Bc%7D%5Ccdot%20%5Chat%20%5Csigma%20%5Csqrt%7B%5Cdfrac%7B1%7D%7Bn_%7B1%7D%7D%2B%5Cdfrac%7B1%7D%7Bn_%7B2%7D%7D%7D%3C%20%5Cmu%20_%7B1%7D-%5Cmu%20_%7B2%7D%3C%20%5Cleft%20%28%5Cbar%7Bx%7D_1-%5Cbar%7Bx%7D_%7B2%7D%20%20%5Cright%20%29%20%2B%20t_%7Bc%7D%5Ccdot%20%5Chat%20%5Csigma%20%5Csqrt%7B%5Cdfrac%7B1%7D%7Bn_%7B1%7D%7D%2B%5Cdfrac%7B1%7D%7Bn_%7B2%7D%7D%7D)
df = n₁ + n₂ - 2
∴ df = 10 + 18 - 2 = 26
From the t-table, we have, for two tails,
= 1.706
![\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}](https://tex.z-dn.net/?f=%5Chat%7B%5Csigma%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Cleft%20%28%20n_%7B1%7D-1%20%5Cright%20%29%5Ccdot%20s_%7B1%7D%5E%7B2%7D%20%2B%5Cleft%20%28%20n_%7B2%7D-1%20%5Cright%20%29%5Ccdot%20s_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B1%7D%2Bn_%7B2%7D-2%7D%7D)
![\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469](https://tex.z-dn.net/?f=%5Chat%7B%5Csigma%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Cleft%20%28%2010-1%20%5Cright%20%29%5Ccdot%20148%5E%7B2%7D%20%2B%5Cleft%20%28%2018-1%20%5Cright%20%29%5Ccdot%20192%5E%7B2%7D%7D%7B10%2B18-2%7D%7D%3D%20178.004321469)
≈ 178
Therefore, we get;
![\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}](https://tex.z-dn.net/?f=%5Cleft%20%281%2C880-1%2C711%20%20%5Cright%20%29%20-%201.706%5Ctimes178%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B10%7D%2B%5Cdfrac%7B1%7D%7B18%7D%7D%3C%20%5Cmu%20_%7B1%7D-%5Cmu%20_%7B2%7D%3C%20%5Cleft%20%281%2C880-1%2C711%20%20%5Cright%20%29%20%2B%201.706%5Ctimes178%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B10%7D%2B%5Cdfrac%7B1%7D%7B18%7D%7D)
Which gives;
![169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}](https://tex.z-dn.net/?f=169%20-%20%5Cdfrac%7B75917%5Ccdot%20%5Csqrt%7B35%7D%20%7D%7B3%2C750%7D%20%3C%20%5Cmu%20_%7B1%7D-%5Cmu%20_%7B2%7D%3C%20169%20%2B%20%5Cdfrac%7B75917%5Ccdot%20%5Csqrt%7B35%7D%20%7D%7B3%2C750%7D)
Therefore, by rounding to the nearest integer, we have;
The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289
Answer:
46 cm
Step-by-step explanation:
Let p represent the length in cm of 1 bap'ai; let k represent the length in cm of 1 bok'ai. Then we have ...
12p +2k = 100
10p +10k = 100
Subtracting the second equation from 5 times the first, we get ...
5(12p +2k) -(10p +10k) = 5(100) -(100)
50p = 400
p = 8 . . . . cm
Then the second equation tells us ...
10(8) +10k = 100
10k = 20
k = 2 . . . . cm
Then 5p+3k = 5(8) +3(2) = 46 cm.
The distance 5 bap'ai and 3 bok'ai is 46 cm.
Answer: the answer to question 6 is order least to greatest
Step-by-step explanation:
Answer:19
Step-by-step explanation:
Triangle=180