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ANEK [815]
3 years ago
7

Bryce lives 26 miles from his grandmother's house .how many kilometers does bryce live from his grandmother's house? round your

answer to the nearest tenth of a mile. 1km 0.62 mi
Mathematics
2 answers:
faltersainse [42]3 years ago
8 0
He lives roughly 41.8 kilometers away from his grandma's house
charle [14.2K]3 years ago
4 0

Answer:

D.The fight to uphold the Petition of Right led to the king's execution and the abolishment of the monarchy

Explanation:

The petition of right was a document written by the english parliment and sent to Charles I because of his bad desicions financially for the country, the petiton of rights demanded that rising taxes the crown took had to be approved by the parliment, also citizens had to be trialed fairly, Charles I signed and soon after was violated by the same King this eventually lead to his execution.

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The radius of a cylinder is 3 cm and the height is 6 cm.
Alborosie

Answer:

B. 54π

Step-by-step explanation:

Surface area of cylinder = 2πrh + 2πr²

Radius = 3 cm

Height = 6 cm

Plug in the values

Surface area of the cylinder = 2*π*3*6 + 2*π*3²

Surface area = 36π + 18π

Surface area = 54π

3 0
3 years ago
10 points!! Will mark brainliest!!! Please help me I don’t understand this very well
Katen [24]

Answer:

9?

Step-by-step explanation:

4 0
3 years ago
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Write what kind of quadrilateral you see in each figure and determine exactly what kind of parallelogram, trapezoid, or trapezoi
irinina [24]

Answer:

Step-by-step explanation:

66.

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7 0
3 years ago
The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo
dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

\to f_X (x) \ \ xe^{-x} \ , \ x>0

For point a:

Moment generating function of X=?

Using formula:

\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx

M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx

integrating the values by parts:

u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0}  -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\  

        = \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\

Therefore, the moment value generating by the function is =\frac{1}{(t-1)^2}

In point b:

E(X^n)=?

Using formula: E(X^n)= M^{n}_{X}(0)

form point (a):

\to M_{X}(t)=\frac{1}{(t-1)^2}

Differentiating the value with respect of t

M'_{X}(t)=\frac{-2}{(t-1)^3}

when t=0

M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\

M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\

7 0
3 years ago
Hey can you please help me posted picture of question
GREYUIT [131]
To solve this problem you must apply the proccedure shown below:

 1. You have the following expression:

 (3+3i)-(13+15i)

 2. If you want to substract both terms, you  need to substract the real numbers and the complex numbers. Then, you obtain:

 3+3i-13-15i
 (3-13)+(3i-15i)

 3. Then, you obtain the following result:

 -10-12i

 4. Therefore, as you can see, the correct answer is the last option (option D), which is:
 D. -10-12i
4 0
4 years ago
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