Solveit

\(A_{rectangle}=40\times20=800\;ft^2\)

5*2+4=14

\(\frac{20}x=\frac{20.25}{50.5}\\ x = 49.8765\)

A: When a baby is​ born, it is a girl.

​B: When a 7​-sided die is​ rolled, the outcome is 4.

Find​ P(A and​ B), the probability that events A and B both occur.

A: a baby can be girl or boy the probability getting one of them is \(\frac12\)

B: on the 7 sided die 7 numbers so the probability getting 4 is \(\frac17\)

Answer:\(\frac12\times\frac17=\frac1{14}\)

\(1+1=2\)

 where is the table ? :/

there s nothing to do with moderators it is just a glutch on the website

\(\left(2.7\cdot\:1017\right)+\left(4.2\cdot\:1015\right)\\ \mathrm{Remove\:parentheses}:\quad\left(a\right)=a\\ =2.7\cdot\:1017+4.2\cdot\:1015\\ \mathrm{Multiply\:the\:numbers:}\:2.7\cdot\:1017=2745.9\\ =2745.9+1015\cdot\:4.2\\ \mathrm{Multiply\:the\:numbers:}\:4.2\cdot\:1015=4263\\ =2745.9+4263\\ \mathrm{Add\:the\:numbers:}\:2745.9+4263=7008.9\\ =7008.9\)