Which polygon has perpendicular sides?

Orlov [11]

a fraction (such as ¹/₂ or ³/₄) in which the numerator and denominator are both integers and are separated by a horizontal or slanted line — compare decimal fraction.

8 0

2 years ago

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There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

2 years ago

Question and answers in imageeeeee

vfiekz [6]

Answer:

red= circumfrence

black= diameter

green= radius

yellow= area

Step-by-step explanation:

3 0

2 years ago

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Plz help nothing to big but hard for me

il63 [147K]

Answer: Which question?

Step-by-step explanation:

5 0

3 years ago

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$