If you are asking for the answer the answer is 0.67.
<span>Volume = L * W * H cubic units
Surface Area = 2 * (LW + LH + WH) square units</span>
A) The solutions to this set of equation is where the graphs cross. They cross at point (-3, -2).
B) The solutions for f(x) would be points that fall on the graph of f(x). Two possible points are (-3, -2) and (-7, 3)
C) These 2 functions cross at (4, 1). That is the solution.
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<h3>a) Box of 10: </h3><h3> up = m ÷ ti </h3><h3> = $4.76 ÷ 10 Cookies</h3><h3> = $0.476 ~ $0.48</h3><h3 /><h3> Box of 25: </h3><h3> up = m ÷ ti </h3><h3> = $10 ÷ 25 Cookies</h3><h3> = $0.4</h3><h3 /><h3>b) Box of 10 - Box of 25</h3><h3> = $0.48 - $0.4</h3><h3> = $0.08</h3><h3 /><h3>THE BOX OF 25 COOKIES IS CHEAPER BY $0.08 PER COOKIE.</h3>
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Answer:
![z= \frac{p- \mu_p}{\sigma_p}](https://tex.z-dn.net/?f=%20z%3D%20%5Cfrac%7Bp-%20%5Cmu_p%7D%7B%5Csigma_p%7D)
And the z score for 0.4 is
![z = \frac{0.4-0.4}{\sigma_p} = 0](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.4-0.4%7D%7B%5Csigma_p%7D%20%3D%200)
And then the probability desired would be:
![P(p](https://tex.z-dn.net/?f=%20P%28p%3C0.4%29%20%3D%20p%28z%3C0%29%20%3D0.5)
Step-by-step explanation:
The normal approximation for this case is satisfied since the value for p is near to 0.5 and the sample size is large enough, and we have:
![np = 45*0.4= 18 >10](https://tex.z-dn.net/?f=%20np%20%3D%2045%2A0.4%3D%2018%20%3E10)
![n(1-p) = 45*0.6= 27 >10](https://tex.z-dn.net/?f=%20n%281-p%29%20%3D%2045%2A0.6%3D%2027%20%3E10)
For this case we can assume that the population proportion have the following distribution
Where:
![\mu_{p}= \hat p = 0.4](https://tex.z-dn.net/?f=%5Cmu_%7Bp%7D%3D%20%5Chat%20p%20%3D%200.4)
![\sigma_p = \sqrt{\frac{p(1-p)}(n} =\sqrt{\frac{0.4(1-0.4)}(45}= 0.0703](https://tex.z-dn.net/?f=%5Csigma_p%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%28n%7D%20%3D%5Csqrt%7B%5Cfrac%7B0.4%281-0.4%29%7D%2845%7D%3D%200.0703)
And we want to find this probability:
![P(p](https://tex.z-dn.net/?f=%20P%28p%20%3C0.4%29)
And we can use the z score formula given by:
![z= \frac{p- \mu_p}{\sigma_p}](https://tex.z-dn.net/?f=%20z%3D%20%5Cfrac%7Bp-%20%5Cmu_p%7D%7B%5Csigma_p%7D)
And the z score for 0.4 is
![z = \frac{0.4-0.4}{\sigma_p} = 0](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.4-0.4%7D%7B%5Csigma_p%7D%20%3D%200)
And then the probability desired would be:
![P(p](https://tex.z-dn.net/?f=%20P%28p%3C0.4%29%20%3D%20p%28z%3C0%29%20%3D0.5)