How to find the missing side of this triangle

In the rectangle below, E1=2x -2, FH=3x+6, and m 4IHG =35°.Find Gl and m LIEH

Tpy6a [65]

Recall that the diagonals of a rectangle bisect each other and are congruent, therefore:

$\begin{gathered} FH=2EI, \\ GI=EI. \end{gathered}$

Substituting the given expression for each segment in the first equation, we get:

$3x+6=2(2x-2).$

Solving the above equation for x, we get:

$\begin{gathered} 3x+6=4x-4, \\ 3x+6+4=4x, \\ 3x+10=4x, \\ 4x-3x=10, \\ x=10. \end{gathered}$

Substituting x=10 in the equation for segment EI, we get:

$EI=2*10-2=20-2=18.$

Therefore:

$GI=18.$

Now, to determine the measure of angle IEH, we notice that:

$\Delta HFG\cong\Delta GEH,$

therefore,

$\measuredangle GHF\cong\measuredangle HGE.$

Using the facts that the triangles are right triangles and that the interior angles of a triangle add up to 180° we get:

$m\measuredangle IEH=90^{\circ}-35^{\circ}=55^{\circ}.$

<h2>Answer: </h2> $\begin{gathered} m\operatorname{\measuredangle}IEH=55^{\operatorname{\circ}}, \\ GI=18. \\ \end{gathered}$

4 0

2 years ago

Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De

Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is, $\lambda=\frac{1}{\mu}=\frac{1}{7}$ .

The probability density function of <em>X</em> is:

$f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0$

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

$P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx$

$=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148$