16 x 8 because its 16 oz per 1 lb so in order to find 8 lbs you have to multiply
Answer:
m∠A= 115
Step-by-step explanation:
Answer:
Explained below.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
![\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
A random sample of <em>n</em> = 658 items is sampled randomly from this population.
As the sample size is large, i.e. <em>n</em> = 658 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample proportion by a Normal distribution.
Compute the mean and standard deviation as follows:
![\mu_{\hat p}=0.61\\\\\sigma_{\hat p}=\sqrt{\frac{0.61(1-0.61)}{658}}=0.019](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%3D0.61%5C%5C%5C%5C%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7B0.61%281-0.61%29%7D%7B658%7D%7D%3D0.019)
(a)
Compute the probability that the sample proportion is greater than 0.63 as follows:
![P(\hat p>0.63)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.63-0.61}{0.019})\\\\=P(Z>1.05)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Chat%20p%3E0.63%29%3DP%28%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D%3E%5Cfrac%7B0.63-0.61%7D%7B0.019%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.05%29%5C%5C%5C%5C%3D1-P%28Z%3C1.05%29%5C%5C%5C%5C%3D1-0.85314%5C%5C%5C%5C%3D0.14686%5C%5C%5C%5C%5Capprox%200.1469)
(b)
Compute the probability that the sample proportion is between 0.60 and 0.66 as follows:
![P(0.60](https://tex.z-dn.net/?f=P%280.60%3C%5Chat%20p%3C0.66%29%3DP%28%5Cfrac%7B0.60-0.61%7D%7B0.019%7D%3C%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D%3C%5Cfrac%7B0.66-0.61%7D%7B0.019%7D%29%5C%5C%5C%5C%3DP%28-0.53%3CZ%3C2.63%29%5C%5C%5C%5C%3DP%28Z%3C2.63%29-P%28Z%3C-0.53%29%5C%5C%5C%5C%3D0.99573-0.29806%5C%5C%5C%5C%3D0.69767%5C%5C%5C%5C%5Capprox%200.6977)
(c)
Compute the probability that the sample proportion is greater than 0.592 as follows:
![P(\hat p>0.592)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.592-0.61}{0.019})\\\\=P(Z>-0.95)\\\\=P(Z](https://tex.z-dn.net/?f=P%28%5Chat%20p%3E0.592%29%3DP%28%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D%3E%5Cfrac%7B0.592-0.61%7D%7B0.019%7D%29%5C%5C%5C%5C%3DP%28Z%3E-0.95%29%5C%5C%5C%5C%3DP%28Z%3C0.95%29%5C%5C%5C%5C%3D0.82894%5C%5C%5C%5C%5Capprox%200.8289)
(d)
Compute the probability that the sample proportion is between 0.57 and 0.60 as follows:
![P(0.57](https://tex.z-dn.net/?f=P%280.57%3C%5Chat%20p%3C0.60%29%3DP%28%5Cfrac%7B0.57-0.61%7D%7B0.019%7D%3C%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D%3C%5Cfrac%7B0.60-0.61%7D%7B0.019%7D%29%5C%5C%5C%5C%3DP%28-2.11%3CZ%3C-0.53%29%5C%5C%5C%5C%3DP%28Z%3C-0.53%29-P%28Z%3C-2.11%29%5C%5C%5C%5C%3D0.29806-0.01743%5C%5C%5C%5C%3D0.28063%5C%5C%5C%5C%5Capprox%200.2806)
(e)
Compute the probability that the sample proportion is less than 0.51 as follows:
![P(\hat p](https://tex.z-dn.net/?f=P%28%5Chat%20p%3C0.51%29%3DP%28%5Cfrac%7B%5Chat%20p-%5Cmu_%7B%5Chat%20p%7D%7D%7B%5Csigma_%7B%5Chat%20p%7D%7D%3C%5Cfrac%7B0.51-0.61%7D%7B0.019%7D%29%5C%5C%5C%5C%3DP%28Z%3C-5.26%29%5C%5C%5C%5C%3D0)
Answer:
There is 16 ounces in one pound.
52/16 = 3.25 lbs = 3lbs 4oz
Step-by-step explanation:
No, because there is not enough information given to make a conclusion. Providing a percentage of 46% higher profit per acre doesn't mean statistical significance as it could be data-based only. There should be a given p value to make a conclusion about statistical significance.