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3 years ago

6

Are the two events “choosing a sophomore” and “choosing someone who replied ‘Yes’” independent events?

1 answer:

4vir4ik [10]3 years ago

5 0

Yes, they are independent event such that occurring the first event “choosing a sophomore” does not affect the probability of occurring the second event <span>“choosing someone who replied ‘Yes’”</span><span />

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I added a picture of the question.

tekilochka [14]

Answer:

C) 2*18, 3*12, 4*9, 6*6

Step-by-step explanation:

When these number sets are multiplied, all of them equal 36, therefore C is your answer.

Hope this helped! :)

7 0

2 years ago

What two formulas are necessary to find the area of a segment of a circle?

Andrew [12]

Answer:

Step-by-step explanation:

Area of a Segment in Radians A = (½) × r2 (θ – Sin θ)

Area of a Segment in Degrees A = (½) × r 2 × [(π/180) θ – sin θ]

3 0

2 years ago

36:38<br> Write the ratio as a fraction in simplest form.

Taya2010 [7]

Answer:

it will 8:9 hope it helps

7 0

3 years ago

I suck at graphs HELP ASAP QUESTION BELOW

kakasveta [241]

ANSWER

A (-∞,-2)

B (0,4).

EXPLANATION

The portion of the graph that is above the x-axis is considered positive.

From the graph the curve is above the x-axis on the interval

(-∞,-2) and (0,4).

The first and second options are correct.

7 0

2 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

2 years ago