Let sample in Texas be p1, Louisiana be p2.
Null Hypothesis:
p1 - p2 = 0
Alternative Hypothesis: (This represents the claim)
p1- p2 < 0
To calculate test statistic, you need the pooled estimate which is a weighted average of the 2 sample proportions.
![P = \frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac{2000(.179+.265)}{2000+2000} = 0.222](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7Bn_1%20p_1%20%2B%20n_2%20p_2%7D%7Bn_1%20%2B%20n_2%7D%20%3D%20%5Cfrac%7B2000%28.179%2B.265%29%7D%7B2000%2B2000%7D%20%3D%200.222)
Next get standard error:
![SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{.222(1-.222)(\frac{1}{2000}+\frac{1}{2000})} = 0.01314](https://tex.z-dn.net/?f=SE%20%3D%20%5Csqrt%7BP%281-P%29%28%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%7D%29%7D%20%3D%20%5Csqrt%7B.222%281-.222%29%28%5Cfrac%7B1%7D%7B2000%7D%2B%5Cfrac%7B1%7D%7B2000%7D%29%7D%20%3D%200.01314)
Calculate test statistic:
![Z = \frac{p_1 - p_2}{SE} = \frac{.179 - .256}{.01314} = -5.86](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bp_1%20-%20p_2%7D%7BSE%7D%20%3D%20%5Cfrac%7B.179%20-%20.256%7D%7B.01314%7D%20%3D%20-5.86)
To find p-value, look up Z-value in standard normal table.
Anything smaller than -3 or larger than 3, you can estimate to have
p-value = 0.
If p-value < alpha, Reject Null Hypothesis.
For this example, 0 < 0.05, therefore reject null hypothesis.
There is evidence to support claim that proportion of smokers in Texas is LESS than Louisiana.
Factor x^3+8
x^3+8
=(x+2)(x^2−2x+4)
Answer:
(x+2)(x^2−2x+4)
Hope that helps!! :) Brainliest??
8 is 4x-24y
10 is c/2-4
Hope that helps