Which expression is equivalent? Please see pictures attached!

The measures of the angles of a convex polygon form an arithmetic sequence. The least measurement in the sequence is 128º. The g

Over [174]

$\bf \qquad \qquad \textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\ a_n=n^{th}~value\\
----------\\
\stackrel{least}{a_1}=128\\\\
\stackrel{greatest}{a_n}=172
\end{cases}$

$\bf S_n=\cfrac{n}{2}(128~+~172)\implies S_n=\cfrac{n}{2}(300)\implies S_n=150n
\\\\\\
\textit{but we also know that sum of all interior angles is }180(n-2)
\\\\\\
\stackrel{\textit{sum of the angles}}{S_n}=\stackrel{\textit{sum of the angles}}{180(n-2)}=150n
\\\\\\
180n-360=150n\implies 30n=360\implies n=\cfrac{360}{30}\implies n=12$

7 0

3 years ago

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The expression 3(x-9) is equivalent to 3(x)+9 3(x)+3(9) 3(x)-9 3(x)-3(9)

ollegr [7]

3(x-9)

when you distribute the equation you get

3(x)-3(9) which is the last answer

6 0

3 years ago

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Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

A rock is thrown up from the ground at a velocity of 84 feet per second. The formula h = -16t2 + 84t gives the rock's height in

kenny6666 [7]

Answer:

h = 110.24 m

Step-by-step explanation:

It is given that,

A rock is thrown up from the ground at a velocity of 84 feet per second. The formula $h = -16t^2 + 84t$ gives the rock's height in feet after t second.