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4 years ago

I’m confused on what’s going on here, so if someone could explain it I would be super grateful.

Mathematics

1 answer:

pshichka [43]4 years ago

8 0

<h2>Range of x </h2><h3> $-\frac{1}{5}$ and x>6</h3><h3>Step-by-step explanation:</h3>

Given AB = 4x + 5, BC = 3x - 2 and AC 9x -5

The sum of the length of any two sides of a triangle is greater than the length of third side.

Case I

AB + BC > AC

⇔4x +5 +3x -2 > 9x -5

⇔7x + 3 >9x - 5

⇔ 7x -9x > -5 -3

⇔-2x > -8

⇔x <4

Case II

if AB+AC > BC

4x +5 +9x -5> 3x -2

⇔13x >3x -2

⇔ 13x -3x> -2

⇔10x > -2

⇔x > $-\frac{1}{5}$

Case III

AC+BC > AB

⇔9x - 5 +3x - 2 > 4x +5

⇔6x -7 > 4x +5

⇔6x -4x >5+7

⇔2x > 12

⇔x > 6

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Random variables X Poisson~ ( a) ,Y Poisson ~ ( a) . X and Y are independent. If 2 1, 2 1. U =2X+ Y-1, V=2X- Y +1. Find: ) Cov (

german

By definition of covariance,

$\mathrm{Cov}(U,V)=E[(U-E[U])(V-E[V])]=E[UV-E[U]V-UE[V]+E[U]E[V]]=E[UV]-E[U]E[V]$

Since $U=2X+Y-1$ and $V=2X-Y+1$ , we have

$E[U]=2E[X]+E[Y]-1$

$E[V]=2E[X]-E[Y]+1$

$\implies E[U]E[V]=(2E[X]+E[Y]-1)(2E[X]-(E[Y]-1))=4E[X]^2-(E[Y]-1)^2=4E[X]^2-E[Y]^2+2E[Y]-1$

and

$UV=(2X+Y-1)(2X-(Y-1))=4X^2-(Y-1)^2=4X^2-Y^2+2Y-1$

$\implies E[UV]=4E[X^2]-E[Y^2]+2E[Y]-1$

Putting everything together, we have

$\mathrm{Cov}(U,V)=(4E[X^2]-E[Y^2]+2E[Y]-1)-(4E[X]^2-E[Y]^2+2E[Y]-1)$

$\mathrm{Cov}(U,V)=4(E[X^2]-E[X]^2)-(E[Y^2]-E[Y]^2)$

$\mathrm{Cov}(U,V)=4V[X]-V[Y]=4a-a=\boxed{3a}$

7 0

4 years ago

Who knows how to do this if u know this help me

skelet666 [1.2K]

Answer:

ITS A AND 2,4,and 1

Step-by-step explanation:

6 0

3 years ago

n the first round of a card game, Niki scored 20 points. Then, she lost 10 points on one turn and an additional 25 points on her

e-lub [12.9K]

The answer would be D, she went from 0 to 20, then lost 10 points which would bring her 10 points, and then she lost an additional 25 points, bringing her to negative 15 (-15)

4 0

3 years ago

Read 2 more answers

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for </u> $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

<em>The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.</em>

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3 years ago

Dewey's soccer team purchased uniforms and equipments for a total cost of $912 . The equipment cost $ 612 , and the uniforms cos

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Answer:

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Step-by-step explanation:

first you subtract 612 from 912 and get 300

then you divide 300 by the cost of uniforms (25)

then you get an answer of 12

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3 years ago

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