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swat32
2 years ago
6

Please Full Explanation Please Will mark Brainlist :))

Mathematics
1 answer:
Anna [14]2 years ago
4 0

Answer:

≈ 3.16 units

Step-by-step explanation:

To determine the distance between the two points, we need to use the distance formula. The distance formula states that;

\sqrt{(x_{2} - x_{1})^{2} + (y_2 - y_1)^{2}  }

Given coordinates;

  • (6, -5)
  • (5, -8)

When we substitute the coordinates into the formula, we get;

  • \implies \sqrt{[5 - 6]^{2} + [-8 - (-5)]^{2}  }

When we simplify the root, we get;

  • \implies \sqrt{[-1]^{2} + [-8 + 5]^{2}  }
  • \implies \sqrt{[-1]^{2} + [-3]^{2}  }
  • \implies \sqrt{1 +9 }
  • \implies \sqrt{10} units ≈ 3.16 units (Using calculator)

Therefore, the distance between the points is about 3.16 units.

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i am planning to attend a company-paid conference. My charges will include $412.00 for airfare, $ 378.00 for conference registra
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Answer:

The estimated expense would be about $1000.00 in which taxzation and inflation is unaccounted for.

Step-by-step explanation:

3 0
3 years ago
An angle whose measure is 90° can be classified as which type of angle?
Lena [83]
Right angle is the right answer
6 0
3 years ago
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A book is in the shape of a right rectangular prism. It measures 8 1/2 inches tall, 7 inches wide, and 2 1/2 inches thick. What
Alenkasestr [34]
Hello!

The formula for volume of a rectangular prism is:

V = lwh

Let's convert the dimensions to decimals, and then use the formula to solve.

8 1/2 = 8.5 as a decimal.

7 = 7 as a decimal.

2 1/2 = 2.5 as a decimal.

Multiply:

V = 8.5 × 7 × 2.5

V = 59.5 × 2.5

V = 148.75

148.75 = 148 3/4 as a fraction because if you convert 3/4 to a decimal by dividing the numerator by the denominator, it equals 0.75.

ANSWER:

The volume of the book is 148 3/4 inches cubed.

6 0
3 years ago
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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

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