9.95^2 x 29.8 = 2,950.2745 and Yes the answer is sensible.
<span>a.
</span>Do you
have sufficient funds to estimate the population mean for the attribute of
interest with a 95% confidence interval 4 units width? Assume that sd= 12
n= {[(Zalpha/2)^2]*[sd]^2}/
se^2
n=
(1.96)^2*(12)^2/ (2)^2
n=
138.297 rounded up to 139
<span>There
is not enough funds for this one
since you’ll need 139 pieces and it costs 10 a piece, you’ll need 1390.</span>
b.
90% confidence interval
n= {[(Zalpha/2)^2]*[sd]^2}/
se^2
n=
(1.645)^2*(12)^2/ (2)^2
n=98
There is enough
funds since 98 pieces for 10 a piece is equal to 980.
Answer:
Option D
Step-by-step explanation:
Given expression has been given as,
![\sqrt[5]{224x^{11}y^8}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B224x%5E%7B11%7Dy%5E8%7D)
![\sqrt[5]{224x^{11}y^8}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 7(x^{11})(y^8)}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B224x%5E%7B11%7Dy%5E8%7D%3D%5Csqrt%5B5%5D%7B2%5Ctimes%202%5Ctimes%202%5Ctimes%202%5Ctimes%202%5Ctimes%207%28x%5E%7B11%7D%29%28y%5E8%29%7D)
![=\sqrt[5]{(2^5)\times (7)(x^{10}\times x)(y^5\times y^3)}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B5%5D%7B%282%5E5%29%5Ctimes%20%287%29%28x%5E%7B10%7D%5Ctimes%20x%29%28y%5E5%5Ctimes%20y%5E3%29%7D)
![=2x^2y\sqrt[5]{7xy^3}](https://tex.z-dn.net/?f=%3D2x%5E2y%5Csqrt%5B5%5D%7B7xy%5E3%7D)
Option D will be the answer.
Answer:
The answer is A x=-6
Step-by-step explanation:
2/3 times -6 = -4.
-4+5=1
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Answer:</u></em></h2><h2><em><u>
Sorry i didn't understand the question but thank you for the points </u></em></h2><h2><em><u>
Step-by-step explanation:</u></em></h2>
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