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3 years ago

Construct line AB 11cm and construct a perpendicular P from line AB 6cm

Mathematics

1 answer:

Arisa [49]3 years ago

8 0

<h3>Step-by-step explanation:</h3>

let us drawn side AB11 cm

again let us drawn p from line AB 6 cm

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= 104.72 square units

Area of the other sector = 2 * 104.72 = 209.44 sq units

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3 years ago

the center of a circle is on the line y=2x and the line x=1 is tangent to the circle at (1,6).find the center and the radius

Julli [10]

The radius and the center of the circle are 4 units and (1,2), respectively

<h3>How to determine the center and the radius?</h3>

The center of the circle is on

y = 2x and x = 1

Substitute x = 1 in y = 2x

y = 2 * 1

Evaluate

y = 2

This means that the center is

Center = (1, 2)

Also, we have the point of tangency to be:

(x, y) = (1, 6)

This point and the center have the same x-coordinate.

So, the distance between this point and the center is

d = 6 - 2

d = 4

This represents the radius

Hence, the radius and the center of the circle are 4 units and (1,2), respectively

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2 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

Construct line AB 11cm and construct a perpendicular P from line AB 6cm​

Construct line AB 11cm and construct a perpendicular P from line AB 6cm