Answer:
The binomial in expanded form is [tex](0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}[/tex].
Step-by-step explanation:
The Binomial Theorem states that a binomial of the form [tex](a + b)^{n}[/tex] can be expanded by using the following identity:
[tex](a + b)^{n} = \Sigma \limits^{n}_{k = 0}\,\frac{n!}{k!\cdot (n-k)!}\cdot a^{n-k}\cdot b^{k}[/tex] (1)
If we know that [tex]a = p = 0.3[/tex] and [tex]n = 5[/tex], then the expanded form of the binomial is:
[tex](p+q)^{n} = \frac{243}{100000} + 5\cdot \left(\frac{81}{10000} \right)\cdot q + 10\cdot \left(\frac{27}{1000})\cdot q^{2} + 10\cdot \left(\frac{9}{100} \right)\cdot q^{3} + 5\cdot \left(\frac{3}{10} \right)\cdot q^{4} + q^{5}[/tex]
[tex](0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}[/tex]
