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wolverine [178]

2 years ago

6

Can anyone please help me answer the PICTURE IS PROVIDED!

1 answer:

kvv77 [185]2 years ago

3 0

Answer is 1 and 3. You’re welcome.

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2 years ago

Please help me find the total area of the composite figure below (geometry)

Akimi4 [234]

Answer:

lw + $\frac{1}{2}$ × π × $(\frac{l}{2} )^{2}$ ⇒ Answer D is correct

Step-by-step explanation:

First, let us find the area of the semi-circle.

Area = $\frac{1}{2}$ × π × r²

<u>Given that,</u>

diameter of the semi-circle is ⇒ <em>l</em>

∴ radius ⇒ <em>l / 2</em>

<u>Let us find it now.</u>

Area = $\frac{1}{2}$ × π × r²

Area = $\frac{1}{2}$ × π × $(\frac{l}{2} )^{2}$

<u> </u>

Secondly, let us find the area of the rectangle.

Area = length × width

<u>Given that,</u>

length ⇒ <em>l</em>

width ⇒ w

<u>Let us find it now.</u>

Area = length × width

Area = l ×w

Area = lw

<u> </u>

And now let us <u>find the total area.</u>

Total area = Area of the rectangle + Area of the semi - circle

Tota area = lw + $\frac{1}{2}$ × π × $(\frac{l}{2} )^{2}$

8 0

1 year ago

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