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Svetach [21]
3 years ago
5

A flea weighs 8 decigrams, and an ant weighs 3 milligrams. How much more does a flea weigh than an ant?

Mathematics
2 answers:
Flauer [41]3 years ago
8 0
1 decigram is 100 milligrams so the flea weighs 800 milligrams and 800-3=797 so the flea weighs 727 milligrams more than the ant
BlackZzzverrR [31]3 years ago
7 0
797
i hope it helped i did for me

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larisa86 [58]
<h3><u>given</u><u>:</u></h3>

<u>\frac{1}{7}   \div 3</u>

<h3><u>solution</u><u>:</u></h3>

<u>\frac{1}{7}  \div  \frac{3}{1}</u>

<u>\frac{1 \times 1}{7 \times 3}</u>

<u>=  \frac{1}{21}</u>

5 0
2 years ago
An angle less than 90 degrees ​
OverLord2011 [107]

Answer:

<u>Acute Angle</u>

Step-by-step explanation:

<u>Obtuse Angle</u> is greater than 90 degrees

<u>Right Angle</u> is exactly 90 degrees

<u>Acute Angle</u> is less than 90 degrees

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3 years ago
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Vladimir79 [104]
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8 0
3 years ago
Express 4x^2 − 12x in the form (2x + a)^2 + b.
gladu [14]

Answer: (2x-3)^2-9\\

a=-3

b=-9

Step-by-step explanation:

4x^2-12x=(2x)^2-2*2x*3+9-9\\=(2x-3)^2-9\\

8 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
3 years ago
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