3 years ago

Profit And Loss

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

0.7% percent is the answer

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What is the arc length of an angle of 2π 3 radians formed on the unit circle? A) π 3 B) 2π 3 C) 4π 3 D) 5π 3

mars1129 [50]

We know that
in a unit circle the radius is 1
therefore
[Length of a circumference]=2*pi*r------> 2*pi

if 2pi radians in a circle unit has a length of ------------> 2*pi
2pi/3 radians----------------------------> X
X=2pi/3

the answer is 2pi/3

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3 years ago

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On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb

stiks02 [169]

Y=kx is the equation for direct variation

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3 years ago

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5) Contains (3,-1) and (6,- 13)<br> Help please

zimovet [89]

Answer:

The slope of the line is -12/3

Step-by-step explanation:

remember, rise over run

My I have brainliest please? :)

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3 years ago

In triangles DEF and OPQ, ∠D ≅ ∠O, ∠F ≅ ∠Q, and segment DF ≅ segment OQ. Is this information sufficient to prove triangles DEF a

zalisa [80]

Answer:

We can't prove this with the SAS postulate.

Step-by-step explanation:

SAS means that two sides and the angle between those two sides are equal.

Side Angle Side (SAS)

That's why we can't prove this with the SAS postulate.

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2 years ago

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

Sati [7]

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$ where: ( $\sigma^2 = \frac{1}{3^2/100}$ )

$P(X \geq0.55) \leq 0.22$

6 0

3 years ago