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vova2212 [387]
3 years ago
7

Find the sum of the geometric series:

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0
\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}\\\\
-------------------------------

\bf \sum\limits_{i=1}^{5}~\stackrel{a_1}{375}\stackrel{r}{\left( \frac{-1}{5} \right)}^{i-1}\qquad 
\begin{cases}
a_1=375\\
r=-\frac{1}{5}
\end{cases}\implies S_5=375\left( \cfrac{1-\left(  \frac{-1}{5}\right)^5}{1-\left(  \frac{-1}{5}\right)} \right)


\bf S_5=375\left( \cfrac{1-\left( \frac{-1}{3125} \right)}{1+\frac{1}{5}} \right)\implies S_5=375\left( \cfrac{1+\frac{1}{3125}}{1+\frac{1}{5}} \right)
\\\\\\
S_5=375\left( \cfrac{\frac{3126}{3125}}{\frac{6}{5}} \right)\implies S_5=375\left(\cfrac{3126}{3125}\cdot \cfrac{5}{6}\right)\implies S_5=375\left(\cfrac{521}{625}\right)
\\\\\\
S_5=\cfrac{1563}{5}\implies S_5=312\frac{3}{5}
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Find the exact value of cos 120° in simplest form with a rational<br> denominator.
IRINA_888 [86]

Given:

cos 120°

To find:

The exact value of cos 120° in simplest form with a rational  denominator.

Solution:

We have,

\cos 120^\circ

It can be written as

\cos 120^\circ=\cos (90^\circ+30^\circ)

\cos 120^\circ=-\sin 30^\circ             [\because \cos (90^\circ-\theta)=-\sin \theta]

\cos 120^\circ=-\left(\dfrac{1}{2}\right)             [\because \sin 30^\circ=\dfrac{1}{2}]

\cos 120^\circ=-\dfrac{1}{2}

Therefore, the exact value of cos 120° is -\dfrac{1}{2}.

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3 years ago
According to the graph, what is the factorization of x² - 5x+6?
pychu [463]

Answer:

OD.(x-3)(x-2)

Step-by-step explanation:

we need two numbers (negative ir positive) that multiply to get 6 abd add to get -5.

so its OD

as -3×-2=6

and -3+-2=-5

5 0
2 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

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Answer:

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Step-by-step explanation:

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