answer:Given: {eq}t_{test} = -2.522 quad text{(left-tailed test)} quad text{and} quad n = 9 quad Rightarrow quad df = 9 - 1 = 8 quad text{(degrees of freedom)} {/eq} The P-value represents the probability of observing a test statistic as extreme or more extreme than the observed one, assuming the null hypothesis is true. For a left-tailed test, the P-value is the area to the left of the test statistic. Since the t-table may not provide an exact P-value, we can use a calculator or software, such as Excel, to approximate it. In our case, we'll calculate the P-value using Excel's TDIST function: {eq}P(t > |t_{test}|) = TDIST(2.522, 8, 1) approx 0.0178487043 {/eq} The P-value for this left-tailed test is approximately 0.0178. Therefore, the range of the P-value is between 0.01 and 0.02.

answer:The perimeter P of the polygon is approximately 0.93. The area A of the polygon can be calculated using the formula for the area of a triangle with coordinates: A = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|. Upon calculation, we get A approx 0.01. The interior angles theta_1, theta_2, and theta_3 can be estimated using the formula theta_i = frac{1}{2} cdot arccos{left(frac{text{det}(P_i, P_{i+1})}{|P_i| |P_{i+1}|}right)}, where P_i and P_{i+1} are vectors formed by the consecutive vertices, and text{det} represents the determinant. The calculated angles are approximately theta_1 approx 1.09, theta_2 approx 0.09, and theta_3 approx 1.96. The polygon is classified as 'Simple' since it does not self-intersect and has non-convex vertices due to the small angle theta_2.

answer:For all natural numbers {eq}n{/eq}, we have {eq}0 < frac{1}{nsqrt{n+1}} leq frac{1}{nsqrt{n}}{/eq}. Consider the series of absolute values of the given alternating series: {eq}sum_{n=1}^{infty} left| frac{(-1)^{n-1}}{nsqrt{n+1}} right| = sum_{n=1}^{infty} frac{1}{nsqrt{n+1}} leq sum_{n=1}^{infty} frac{1}{nsqrt{n}} = sum_{n=1}^{infty} frac{1}{n^{3/2}}{/eq}. Since {eq}sum_{n=1}^{infty} frac{1}{n^{3/2}}{/eq} is a p-series with {eq}p = 3/2 > 1{/eq}, it converges by the p-series test. As the series {eq}sum_{n=1}^{infty} frac{1}{nsqrt{n+1}}{/eq} is smaller than a convergent p-series, it is also convergent. Therefore, the original alternating series {eq}sum_{n=1}^{infty} frac{(-1)^n}{nsqrt{n+1}}{/eq} is absolutely convergent and hence convergent.

answer:The inverse of the matrix is: [ left( begin{array}{ccc} -frac{180}{457} & frac{93}{914} & -frac{75}{914} frac{273}{457} & frac{78}{457} & frac{114}{457} -frac{42}{457} & -frac{12}{457} & -frac{123}{457} end{array} right) ]