What is the solution to 34x+95=3(14x +9)

How do you prove theorems behind two parallel lines cut by a transversal?

LenaWriter [7]

The way to do it can be explained like this:
Say AB and CD are the two parallel lines cut by a transversal at E and F respectively.
Then the pairs of alternate interior angles are:
Angle(AEF) and Angle(DFE)
Angle(CFE) and Angle(BEF)
Now lets prove if this is true:
Angle(CFE) +Angle(DFE) = 180
(linear pair)
Also
Angle(CFE) +Angle(AEF) = 180
(Corresponding angles)
Equate the above results:
Angle(CFE) +Angle(DFE) = Angle(CFE) +Angle(AEF)
Angle(DFE) = Angle(AEF)
Happens the same with
Angle(CFE) = Angle(BEF)
Hope this is very useful for you

5 0

3 years ago

How to find average value of a function over a given interval?

Strike441 [17]

f(x)=8x−6f(x)=8x-6 , [0,3][0,3]

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.(−∞,∞)(-∞,∞){x|x∈R}{x|x∈ℝ}f(x)f(x) is continuous on [0,3][0,3].f(x)f(x) is continuousThe average value of function ff over the interval [a,b][a,b] is defined as A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dx.A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dxSubstitute the actual values into the formula for the average value of a function.A(x)=13−0(∫308x−6dx)A(x)=13-0(∫038x-6dx)Since integration is linear, the integral of 8x−68x-6 with respect to xx is ∫308xdx+∫30−6dx∫038xdx+∫03-6dx.A(x)=13−0(∫308xdx+∫30−6dx)A(x)=13-0(∫038xdx+∫03-6dx)Since 88 is constant with respect to xx, the integral of 8x8x with respect to xx is 8∫30xdx8∫03xdx.A(x)=13−0(8∫30xdx+∫30−6dx)A(x)=13-0(8∫03xdx+∫03-6dx)By the Power Rule, the integral of xx with respect to xx is 12x212x2.A(x)=13−0(8(12x2]30)+∫30−6dx)

3 0

4 years ago

A game involves spining this spinner. what is the probality that the pointer will land on g?

Doss [256]

There's a 1 in 7 chance it will land on G.

6 0

3 years ago

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The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag

VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{2050}{\sqrt{49}} = 574$

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0

3 years ago

Combina los términos semejantes -16 + 8x +20x -5y

mrs_skeptik [129]

28x+-21 you simplify them first

7 0

3 years ago

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