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mina [271]
2 years ago
11

Expressions are equivalent. If so, tell what property is applied. 8 × 1 and 8

Mathematics
1 answer:
mars1129 [50]2 years ago
5 0

Answer:

Step-by-step explanation:

Identity property of multiplication

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Answer: 9 cm

Step-by-step explanation:

By multiplying 5 by 3, you get 15. this would also apply to finding the other side of the rectangle, where you multiply 3 by 3 to get 9.

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Stolb23 [73]
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8 0
3 years ago
Quadrilateral ABCD is inscribed in a circle. Find the measure of each of the angles of the quadrilateral.
Nadusha1986 [10]

Here is a reference to the Inscribed Quadrilateral Conjecture it says that opposite angles of an inscribed quadrilateral are supplemental.

Explanation:

The conjecture, #angleA and angleC# allows us to write the following equation:

#angleA + angleC=180^@#

Substitute the equivalent expressions in terms of x:

#x+2+ x-2 = 180^@#

#2x = 180^@#

#x = 90^@#

From this we can compute the measures of all of the angles.

#angleA=92^@#

#angleB=100^@#

#angleC=88^@#

<span>#angleD= 80^@#</span>

4 0
3 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
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
2 years ago
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