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hichkok12 [17]
3 years ago
13

Find the value of x *​

Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

Answer: its 4 lol just look at the other side and you see the 4 there lol

Step-by-step explanation:

You might be interested in
Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive
PtichkaEL [24]
Looks like the PMF is supposed to be

\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65

Next,

\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25

\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5

\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2
\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5
\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}

If Y=X^2+1, then X^2=Y-1\implies X=\sqrt{Y-1}, where we take the positive root because we know X can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that Y can take on the values 1^2+1=2, 2^2+1=5, and 5^2+1=26. At these values of Y, we would have the same probability as we did for the respective value of X. That is,

\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}

Part (5) is incomplete, so I'll stop here.
8 0
3 years ago
Help!! help!! help!! help!! help!! help!! help!! help!! help!!
oksian1 [2.3K]

Answer:

<em>(8.21, -20.79)</em>

Step-by-step explanation:

Given the simultaneous equation;

\frac{a}{b} - \frac{b}{2} = 10\\a - b = 29

From 2;

a = 29 + b ....3

Substitute 3 into 1;

\frac{29+b}{b} - \frac{b}{2}  = 10\\\frac{2(29+b)-b^2}{2b} = 10\\\frac{58+2b-b^2}{2b} = 10\\58+2b-b^2 = 20b\\b^2-2b-58 = -20b\\b^2-2b+20b -58 = 0\\b^2+18b-58 = 0\\

Factorize

b = -18±√18²-4(-58)/2

b = -18±√324+232/2

b = -18±√556/2

b = -18±23.58/2

b = -18-23.58/2 and -18+23.58/2

b = -41.58/2 and 5.58/2

b = -20.79 and 2.79

Since a = 29 + b

when b = -20.79

a = 29 - 20.79

a = 8.21

<em>Hence the solution to the system of equation is (8.21, -20.79)</em>

7 0
2 years ago
How will the graph of line y = -5x + 7 be affected if the 7 is changed to –7?
dybincka [34]

Answer:a

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
If anyone knows about definite integrals for calculus then please I request help! I
kicyunya [14]

Answer:

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [Bounds] Switch:                                                                                           \displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx
  2. [Integral] U-Substitution:                                                                              \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du
  3. [Integral] Exponential Integration:                                                               \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)
  5. Simplify:                                                                                                         \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0
2 years ago
Colby is making a home video consisting of a 5-minute introduction followed by several short skits. Each skit is 6 minutes long.
givi [52]
<h2>Hello!</h2>

The answer is: There are a total of 16 skits in the video.

<h2>Why?</h2>

From the statement, we know that the total length of the video is 101 minutes and it contains a 5-minute introduction, the rest of the minutes are short skits (which are 6 minutes long each one).

So, we need to write the following equations:

First equation:

TotalDuration=Introduction+n*Skit

Second equation:

Skit=6Minutes

Where,

TotalDuration=101Minutes\\Introduction=5Minutes\\n=NumberOfSkits\\Skit=6Minutes

So, replacing and substituting the second equation into the first equation we have:

101Minutes=5Minutes+n*6Minutes\\n*6Minutes=101Minutes-5Minutes\\n*6Minutes=96Minutes\\n=\frac{96Minutes}{6Minutes}=16

So, there are a total of 16 skits in the video.

Let's prove it substituting "n" into the first equation:

101Minutes=5Minutes+16*6minutes\\101Minutes=5Minutes+96Minutes\\101Minutes=101Minutes

Have a nice day!

8 0
3 years ago
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