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svetoff [14.1K]
2 years ago
8

Please help! Math test not much time! :)

Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
4 0
It will take 512 minutes to complete station 8.
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Please help me to solve these questions as soon as possible. Thank you ​
densk [106]

Answer:

499

9

April 12 2021

Step-by-step explanation:

1 and 500 are the smallest and largest

12 is the highest common factor

Every 12 days they have an appointment together

8 0
3 years ago
I don’t feel like doing math
Zinaida [17]

Answer:

26.14

Step-by-step explanation: Just subtract.

7 0
3 years ago
Kim namjoon...<br><br> anyone army??
vova2212 [387]

Answer:

Er... um... huh?

Step-by-step explanation:

3 0
3 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
RoseWind [281]

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
In a certain community, 30% of the families own a dog, and 20% of the families that own a dog also own a cat. It is also known t
Komok [63]

Answer:

What is the probability that a randomly selected family owns a cat? 34%

What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%

Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:

Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)

So, 30% of the families own a dog = .30*100 = 30

20% of the families that own a dog also own a cat = 0.2*30 = 6

34% of all the families own a cat = 0.34*100 = 34

Dogs and cats: 6

Only dogs: 30 - 6 = 24

Only cats: 34 - 6 = 28

Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42

What is the probability that a randomly selected family owns a cat?

34/100 = 34%

What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?

A = doesn't own a dog

B = owns a cat

P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%

5 0
3 years ago
Read 2 more answers
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