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2 years ago

Solve the system of equations by substitution.

Mathematics

1 answer:

n200080 [17]2 years ago

4 0

Answer:

x = -8

y = -6

Step-by-step explanation:

y = -2x - 22

y = x + 2

x + 2 = -2x - 22

x + 2x = -22 - 2

3x = -24

x = -24 : 3

x= -8

y = -8 + 2

y = -6

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The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago

An adventurous dog strays from home, runs three blocks east, two blocks north, one block east, one block north, and two blocks w

Marrrta [24]

<h2>Answer:</h2>

<u>Distance = 360.55 m</u>

<u>Direction = North-East</u>

<h2>Step-by-step explanation:</h2>

In the question,

The adventurous dog starts from O, say.

He moves 3 blocks East = 300 m

then,

2 blocks North = 200 m

1 block East = 100 m

1 block North = 100 m

2 blocks West = 200 m

So,

Distance of the Dog's <u>final position</u> from the <u>initial position</u><u>.</u>

OE is given by,

In triangle OLE, using Pythagoras theorem,

$OE^{2}=OL^{2}+LE^{2}\\OE^{2}=(200)^{2}+(300)^{2}\\OE^{2}=40000+90000\\OE=360.55\,m$

<em><u>Therefore, the distance of Dog from the Home is 360.55 m and the direction of Dog from Home is North-East.</u></em>

8 0

3 years ago

17 __ a solution od d - 6 = 23<br><br> is it or is it not

anygoal [31]

Answer:

29 is the answer.

:-)

;-)

3 0

2 years ago

Read 2 more answers

why does the multiplication property of equality not allow us to divide both sides of an equation by zero

AlladinOne [14]

I’m not sure you can look this up

5 0

3 years ago

Find several points that are on the graph of the function y = x^2 + 1. Plot the points in a coordinate plane.

beks73 [17]

Answer:

Below.

Step-by-step explanation:

Plot following points.

Calculate the point by plugging in values of x into x^2 + 1

for example When x = 0, y = 0^1 + 1 = 1.

So plot plot (0, 1),

Make a table of points to plot:

x -3 -2 -1 0 1 2 3

y 10 5 2 1 2 5 10

When you plot the points you'll see the graph is U shaped.

The function is of second degree (as it contains x^2) so it wont be linear.

8 0

3 years ago