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2 years ago

If m∠9 = 130°, what is m∠4?

Mathematics

1 answer:

Bad White [126]2 years ago

6 0

Answer:

m<9=130 it also equal to 180 so 130-180 makes m<4= 50

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4. Dean Pelton wants to perform calculations to impress the accreditation consultants, but upon asking for information about GPA

Leya [2.2K]

Answer:

the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

Step-by-step explanation:

Let assume that n should represent the number of the students

SO, $\bar x$ can now be the sample mean of number of students in GPA's

To obtain n such that $P( \bar x \leq 2.3 ) \leq .04$

⇒ $P( \bar x \geq 2.3 ) \geq .96$

However ;

$E(x) = \int\limits^4_2 Dx (2+e^{-x} ) 4x = D \\ \\ = D(e^{-x} (e^xx^2 - x-1 ) ) ^D_2 = 12.314 D$

$E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D$

Similarly;

$D\int\limits^4_2(2+ e^{-x}) dx = 1$

⇒ $D*(2x-e^{-x} ) |^4_2 = 1$

⇒ $D*4.117 = 1$

⇒ $D= \dfrac{1}{4.117}$

$\mu = E(x) = 2.991013 ; \\ \\ E(x^2) = 9.28103$

∴ $Var (x) = E(x^2) - E^2(x) \\ \\ = .3348711$

Now; $P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega \geq .691013) \ \ \ \ \ \ \ \ \ \ (x = E(\bar x ) - \mu)$

Using Chebysher one sided inequality ; we have:

$P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}$

So; $(\omega = \bar x - \mu)$

⇒ $E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}$

∴ $P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}$

To determine n; such that ;

$\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}$

⇒ $n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}$

$n \geq 16.83125$

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

5 0

2 years ago

The roof of a tower in a castle is shaped like a cone. The height of the roof is 30 ft, and the radius of the base is 15 ft. Wha

xz_007 [3.2K]

Answer:

The area of the roof: ≈2287.44 ft^2, the lateral area of the roof:≈1580.58 ft^2

Step-by-step explanation:

The area of the roof is computed by the equation of the area of a cone:
A = πr(r + $\sqrt{h^{2}+r^(2) }$ )
(r: radius, h: height, πr^2 is the area of the base of the cone, πr $\sqrt{h^{2}+r^(2) }$ is the lateral area of the cone). So:

A = $15\pi (15 + \sqrt{30^{2} +15^{2} } )$ ≈ 2287.44 ft^2.

As I stated earlier, the lateral area of the roof can be computed:

A $L$ = $15\pi \sqrt{30^{2} +15^{2} }$ ≈ 1580.58 ft^2.

5 0

1 year ago

PLEASE HELPPPPP

AnnyKZ [126]

An even prime number would be 4,8,10

3 0

2 years ago

Please help with questions<br> 1d) and 1e)<br><br> Any help is much appreciated thanks

Anestetic [448]

Answer:

1d 6.742 1e 6

Step-by-step explanation: By sketching out the graph you can see the positive x intercept to find 1d

You can also replace h with 25 and isolate t to find 1e

5 0

2 years ago

Graph (1,1), (1,2), (1,3), (1,4)

jenyasd209 [6]

Answer:

There is not a way that I can show you how to graph it, so I will try and explain it.

Step-by-step explanation:

(1,1)

^ ^ y-axis

x-axis

You are just going right however many, and then up however many it says.

So if it's (1,1) you will go right one space, and then up one space.

Once you have all of your points on the graph, you put a line through them.

Hope this helps! :)

3 0

2 years ago