Find the Area of the Parallelogram.

Mathematics

1 answer:

OLga [1]2 years ago

6 0

Answer:

should be 84

Step-by-step explanation:

use the 6 and 14 and solve as you would a rectangle. if you move the little triangle to the other side of the parallelogram it becomes a rectangle and the problem is solved

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In the local frisbee league, teams have 7 members and each of the 4 teams takes turns hosting tournaments. At each tournament, e

mars1129 [50]

There are 4 teams in total and each team has 7 members. One of the team will be the host team.

Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.

So total number of possible 9 member tournament committees will be equal to:

$7C3 \times 7C2 \times 7C2 \times 7C2\\ \\ = 324135$

This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:

$9 \times 324135\\ \\ =2917215$

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2 years ago

there are 63 students marching in a band, and they'are marching in 7 rows. how many students are in each row?

photoshop1234 [79]

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5 0

3 years ago

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What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$