The function is given
.
Now let's solve for
.
![f(-2)=2^{-2}+3](https://tex.z-dn.net/?f=f%28-2%29%3D2%5E%7B-2%7D%2B3)
Rule of negative exponent:
.
![f(-2)=\frac{1}{2^2}+3](https://tex.z-dn.net/?f=f%28-2%29%3D%5Cfrac%7B1%7D%7B2%5E2%7D%2B3)
![f(-2)=\frac{1}{4}+\frac{3}{1}](https://tex.z-dn.net/?f=f%28-2%29%3D%5Cfrac%7B1%7D%7B4%7D%2B%5Cfrac%7B3%7D%7B1%7D)
Rule for sum of fractions:
.
![f(-2)=\frac{13}{4}](https://tex.z-dn.net/?f=f%28-2%29%3D%5Cfrac%7B13%7D%7B4%7D)
And the result is:
![f(-2)=\boxed{\frac{13}{4}=3.25}](https://tex.z-dn.net/?f=f%28-2%29%3D%5Cboxed%7B%5Cfrac%7B13%7D%7B4%7D%3D3.25%7D)
Answer:
0.0499
Step-by-step explanation:
This is a binomial probability function expressed as:
![P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D%5C%5C%5C%5C)
Given that n =8, and p(male)=1-0.6=0.4, the probability of at least 6 being male is calculated as:
![P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 6)=P(X=6)+P(X=7)+P(X=8)\\\\={8\choose 6}0.4^6(0.6)^{2}+{8\choose 7}0.4^7(0.6)^{1}+{8\choose 8}0.4^8(0.6)^{0}\\\\=0.0413+0.0079+0.0007\\\\=0.0499](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5Ex%281-p%29%5E%7Bn-x%7D%5C%5C%5C%5CP%28X%5Cgeq%206%29%3DP%28X%3D6%29%2BP%28X%3D7%29%2BP%28X%3D8%29%5C%5C%5C%5C%3D%7B8%5Cchoose%206%7D0.4%5E6%280.6%29%5E%7B2%7D%2B%7B8%5Cchoose%207%7D0.4%5E7%280.6%29%5E%7B1%7D%2B%7B8%5Cchoose%208%7D0.4%5E8%280.6%29%5E%7B0%7D%5C%5C%5C%5C%3D0.0413%2B0.0079%2B0.0007%5C%5C%5C%5C%3D0.0499)
Hence, the probability of at least 6 males is 0.0499
425.82 + 120.75
Then subtract withdrawal amount
Is $360.67
the answer is the third one