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3 years ago

12

Solve log2(6-2x)-log2x=3

1 answer:

Elenna [48]3 years ago

6 0

Assuming you have log in base 2.

Join logarithms (always):

log2(6-2x)-log2(x) = log2( (6-2x/x) )

Then get rid of it! :)

(6-2x)/x = 2^3=8,

Solve the equation:

6-2x=8x ---> 10x=6 ---> x = 3/5.

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3 years ago

If 3/4 of a ton of sand covers 2/5 of a playground, how many tons of sand are required to cover the entire playground?

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4 years ago

a flower bed is in the shape of a triangle with one side twice the length of the shortest side and the third side is 19 feet mor

Neporo4naja [7]

Let's assume

shorter side as x

so, first side =2x

third side =x+19

now, we can find perimeter of triangle

perimeter of any triangle is

first side + shortest side + third side

so, we get

$P=2x+x+x+19$

we are given perimeter=187

so, we can set it to 187

and then solve for x

$187=2x+x+x+19$

$187=4x+19$

$4x+19-19=187-19$

$4x=168$

$x=42$

so,

shortest length is x=42 feet

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3 0

4 years ago

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