“We have ground to make up and that’s why we believed it would be like a playoff game. “I think both of us are playing really good ball these last couple of weeks,” Niners coach Kyle Shanahan said. 500 teams fighting for wild-card spots was a highly entertaining one with six scores in the third quarter alone, including a 99-yard kickoff return score by Minnesota’s Kene Nwangwu.īut the 49ers (6-5) made just enough plays, including a fourth-down stop near the goal line midway through the fourth quarter to win their third straight following a stretch of five losses in six games. We had to execute better and now we’re starting to execute at the right time.”
“We’re all excited to be back in the mix and have the season in our control,” defensive end Nick Bosa said. READ MORE: Rain, Snow, Freezing Temperatures On Deck For Northern Californiaĭeebo Samuel ran for two touchdowns before leaving with an injury and Elijah Mitchell rushed for 133 yards and a score, leading the 49ers to a 34-26 victory over the Minnesota Vikings on Sunday. (AP) - The midseason skid that looked as though it could derail the San Francisco 49ers season is firmly in the past.Īfter getting back to a successful formula of dominating on the ground, the Niners have won three straight games and again look like a contender in the NFC. Then you would add up ALL of those throws and divide by $100$ and get $\approx 6.SANTA CLARA, Calif. About $9$ times it will take $4$ throws( $36$ throws) etc. About $11$ times it will take $3$ throws( $33$ throws). About $14$ times it will take $2$ throws ( $28$ throws).
About $17$ times it will only take $1$ throw( $17$ throws). You can imagine an experiment with $100$ trials. It doesn't mean much except to illustrate why the mean number of throws to get the first $6$ is higher than around $3$ or $4.$ And since we're after a weighted average, that is multiplied by $4$ to get the value at $x=4$. There is about a $0.0965$ chance of getting the first roll of a $6$ on the $4$th roll. It's not a great graph, honestly, as it is kind of abstract in what it represents. The probability of rolling your first $6$ on the $n$-th roll is $$\left$ You can calculate the average this way also.
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