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2 years ago

10

The height of an equilateral triangle is 11√12. What is a side of the triangle?

1 answer:

Sonja [21]2 years ago

4 0

Answer:

I don't know the answer

Step-by-step explanation:

it's hard solve it for me

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A child’s wading pool has a diameter of 5 feet and a height of 2 feet. How much water would it take to fill the pool?

Svetradugi [14.3K]

Answer:

12.5π or ≈39.27

Step-by-step explanation:

The formula for finding the volume is V=πr^2*d (where h is the height and r is the radius).

Plug in the values: V=π(2.5)^2*2 (Diameter=2*Radius)

Solve: V=6.25π*2

V=12.5π

V≈39.27

3 0

2 years ago

Look at the graph .which is the relationship between x and y PLSSS HURRY (100 POINTS

skelet666 [1.2K]

Answer:

it is B

Step-by-step explanation:

6 0

2 years ago

Answer has to be in units

chubhunter [2.5K]

It's a deal. Give me a question, and I'll give you an answer in units.

7 0

3 years ago

Read 2 more answers

23 ones x 2 tens= answer

shutvik [7]

23 and 2 tens is 23 and 20 which equals 53

53

5 0

3 years 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

3 years ago