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2 years ago

Around which line would the following cross-section need to be revolved to create a sphere?

Mathematics

2 answers:

umka21 [38]2 years ago

4 0

Y=x is the answer for the coordinates

masya89 [10]2 years ago

3 0

Answer:

rotating around y = 1 will make the sphere.

Step-by-step explanation:

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2 years ago

Two trains are moving towards each other. One train moves at a speed of 50 mph, and the other train travels at 60 mph. How soon

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2 years ago

Read 2 more answers

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

2 years ago

A culture started with 3000 bacteria. After 4 hours, it grew to 3,600 bacteria. Predict how many bacteria will be present after

Vlad [161]

Hello Shannonrodrigue. You can solve this by setting up an exponential growth equation.

$A = A_ob^t$
$3600 = 3000b^4$

Now we solve for b
$3600 = 3000b^4$
$\frac{3600}{3000} = \frac{3000b^{4}}{3000}$
$\frac{6}{5} = b^4$
$\sqrt[4]{\frac{6}{5}} = \sqrt[4]{b^4}$
$\sqrt[4]{\frac{6}{5}} = b$

Now that we have found b, we can use the equation $A = 3000b^t$ to predict how many bacteria will be present after 10 hours.
$b = \sqrt[4]{\frac{6}{5}}$
t = 10
$A = 3000b^t$
$A = 3000\sqrt[4]{\frac{6}{5}}^{10} = 4732.3228 = 4732$

Answer = 4732

4 0

2 years ago