Fill in the blanks with whole numbers. Round up if needed!

A UART frame consists of a start bit that is $,$ a stop bit that is

$$ a parity bit $($optional$),$ and $$ bits of data. For the following assume that we have a parity bit. If you want to transfer $10$ bytes of data, you have to transfer $$ bits. Given a baud rate of $9600bps,$ transferring these bytes would take almost msec.

Using odd parity, the parity bit of the data byte $00111100$ is

Using even parity, the parity bit of the data byte $1110000$ is $$