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Travka [436]
2 years ago
11

What types of Functions will left endpoint lead to an under approximation and right endpoints lead to and over approximation?

Mathematics
1 answer:
zepelin [54]2 years ago
8 0

The functions that the left endpoint lead to an under approximation and right endpoints lead to and over approximation is the positive and increasing function.

<h3>How to illustrate the function?</h3>

It should be noted that function simply means the illustration that shows the relationship between the variables.

In this case, the functions that the left endpoint lead to an under approximation and right endpoints lead to and over approximation is the positive and increasing function.

Learn more about functions on:

brainly.com/question/25638609

#SPJ1

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What is 18 divided by 11
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❗️❗️❗️Find the length of side x in simplest radical form with a rational denominator.❗️❗️❗️
tiny-mole [99]

Answer:

x=\frac{\sqrt{14} }{2}

Step-by-step explanation:

Notice that you are given an isosceles right-angle triangle to solve, since each of its two acute angles measures 45^o. Then such means that the sides opposite to these acute angles (the so called "legs" of this right angle triangle) must also be of the same length (x).

We can then use the Pythagorean theorem that relates the square of the hypotenuse to the addition of the squares of the triangles legs:

(\sqrt{7})^2=x^2+x^2\\7=2\,x^2\\x^2=\frac{7}{2} \\x=+/-\sqrt{\frac{7}{2}} \\x=+/-\frac{\sqrt{14} }{2}

We use just the positive root, since we are looking for an actual length. then, the requested side is:

x=\frac{\sqrt{14} }{2}

6 0
3 years ago
A line passes through the point (10,3)
pishuonlain [190]
We have to find b first:

3 = -3/2(10)+b
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5 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
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
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