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algol13
4 years ago
15

Approximately 10.3% of american high school students drop out of school before graduation. choose 10 students entering high scho

ol at random. find the probability that all 10 stay in school and graduate. round your answer to the nearest thousandth.
Mathematics
1 answer:
Hatshy [7]4 years ago
4 0
The probability that a binomial experiment of n trials and p, the probability of success, produces an outcome, x is given by:

P(x)={ ^nC_r}p^x(1-p)^{n-x}

Given that a<span>pproximately 10.3% of american high school students drop out of school before graduation, thus p = 0.103.

If we choose 10 students (i.e. n = 10)

The </span><span>probability that all 10 stay in school and graduate means the probability that none of the 10 students dropped outP(0)={ ^{10}C_0}(0.103)^0(1-0.103)^{10} \\  \\ 1\times1\times(0.897)^{10}=0.337 is given by


</span>
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Answer:

Step-by-step explanation:

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):
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<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)

First pair of opposite sides:

<u>Slope:</u>

\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}

Second pair of opposite sides:

<u>Slope:</u>

\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\

So the diagonals measure the same, therefore this is a rectangle.

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The graphs of ​ f(x)=2x+4 ​ and ​ g(x)=10−4x ​ intersect at (1, 6) . What is the solution of the equation 2x+4=10−4x ?
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Answer:

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