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lilavasa [31]
2 years ago
5

Please help me graph this inequality

Mathematics
1 answer:
DaniilM [7]2 years ago
7 0

Answer:

rearrange to get x < -4

Step-by-step explanation:

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R, S, Q and P are the midpoints of OC, CB, BA and OA respectively.
Stolb23 [73]

Answer:

See below.

Step-by-step explanation:

Draw segment OB.

In triangle OBC, points R and S are the midpoints of sides OC and BC, respectively. That makes RS parallel to OB.

In triangle OBA, points P and Q are the midpoints of sides OA and BA, respectively. That makes PQ parallel to OB.

Since segments RS and PQ are parallel to segment OB, then RS and PQ are parallel to each other.

3 0
3 years ago
Vanessa's scale drawing of a volleyball court, shown below, is 3 inches long and 1 1/2 inches wide. How many square meters make
natima [27]
I took a test on this yesterday and all i know is the answer is either 81 117 or 162
6 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
My mom ran a errand that took 45 minutes if she left at 2:10 p.m. what time did she return​
IRISSAK [1]

Answer:

2:55 p.m

Step-by-step explanation:

Have a Great Day!!

4 0
3 years ago
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Use the function below to find F(3)<br> F(x) = (1/6)^x
alukav5142 [94]

Answer: 1/216

Step-by-step explanation: please mark me brainly

7 0
2 years ago
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