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3 years ago

Connect A, B, and C to make a closed polygon.

Mathematics

1 answer:

snow_lady [41]3 years ago

6 0

Answer:

c. 4 + 4√2 units

b. four square units

a. AB ___2√2____ BC ___2√2____ AC ___4___

Step-by-step explanation:

c. According to the 45°-45°-90° triangle[s] theorem, where the legs are identical [√2], the diagonals form hypotenuses, so for every diagonal unit you go up, you count √2, also since this an isosceles right triangle. With that being said, you have this:

$\displaystyle 2[2\sqrt{2}] + 4 = 4\sqrt{2} + 4$

b. $\displaystyle \frac{1}{2}hb, \frac{1}{2}bh, or\: \frac{bh}{2} = A$

The height is two units, just so you know.

a. BC and AB are equivalent in length, so they both receive 2√2, then AC is the simple one at four units.

I am joyous to assist you anytime.

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rusak2 [61]

Answer:

a =2.2

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2+b^2 = c^2

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2 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

3 years ago

Connect A, B, and C to make a closed polygon.​

Connect A, B, and C to make a closed polygon.