1918, approximately \( 60 \% \) of the wolves in a region were male, and \( 40 \% \) were female. However, cattle ranchers in egrea have made a determined effort to exterminate wolves. From 1918 to the present, approximately \( 70 \% \) of wolves in \( 30 \% \) are female. Biologists suspect that male wolves are more likely than females to return to an where the population has been greatly reduced. (Round your answers to three decimal places.) Before 1918 , in a random sample of 11 wolves spotted in the region, what is the probability that 8 or more were male? What is the probability that 8 or more were female? \( \square \) What is the probability that fewer than 5 were female? \( \square \)

On considère la suite \( \left(u_{n}\right) \) définie par \( u_{0}=0 \) et, pour tout entier naturel \( n, u_{n+1}=3 u_{n}-2 n+3 \). 1. Calculer \( u_{1} \) et \( u_{2} \). 2. Démontrer par récurrence que, pour tout entier naturel \( n, u_{n} \geqslant n \). 3. En déduire que la suite \( \left(u_{n}\right) \) est croissante. 4. Soit la suite \( \left(v_{n}\right) \) définie, pour tout entier naturel \( n \), par \( v_{n}=u_{n}-n+1 \). a. Démontrer que la suite \( \left(v_{n}\right) \) est une suite géomé- trique de raison 3 . b. En déduire que, pour tout entier naturel \( n \) : \( \quad u_{n}=3^{n}+n-1 \). c. Calculer \( \sum_{k=0}^{100} u_{k}=u_{0}+u_{1}+\ldots+u_{100} \) On rappelle que, pour tout entier naturel \( n \) : \( 1+2+\ldots+n=\frac{n(n+1)}{2} \) et, pour \( q \neq 1 \) : \( \quad 1+q+q^{2}+\ldots+q^{n}=\frac{1-q^{n+1}}{1-q} \).

Using the Intermediate Value Theorem, determine if there exists at least one zero for the polynomial function \( f(x) = x^{3} - 4x + 1 \) in the interval \([-3, 0]\).

1.- Resuelve las siguientes ecuaciones: a) \( 2 \sqrt{4 x+1}+3=6+\sqrt{x+7} \) b) \( \frac{1}{2} \log (8 x+2)-\frac{1}{2} \log (x-4)=1-\log 2 \)

\( \frac{dy}{dx}(x^{2}y^{3}+xy)=1 \)

ponential form. \( \begin{array}{l}\text { 17) } x^{4} \times 3 x= \\ \text { 18) } x \times 2 x^{2}= \\ \text { 19) } 5 x^{4} \times 5 x^{4}= \\ \text { 20) } 2 \mathrm{y} x^{2} \times 2 x= \\ \text { 21) } 3 x^{4} \times y^{2} x^{4}= \\ \text { 22) } y^{2} x^{3} \times y^{5} x^{2}= \\ \text { 23) } 4 y x^{3} \times 2 x^{2} y^{3}= \\ \text { 24) } 6 x^{2} \times 6 x^{3} y^{4}=\end{array} \)