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2 years ago

15

Please help show work I beg A 50 B 13 C 12 D 51

1 answer:

Viefleur [7K]2 years ago

3 0

B. this answer would be B.

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Find the length of side x in simplest radical form with a rational denominator

Angelina_Jolie [31]

Answer:

The length of x=3/1

Step-by-step explanation:

Given the right triangle

Ф = 30

We know that

sin Ф = opposite/hypotenuse

The opposite of angle 30° = x

The hypotenuse is = 6

so

sin 30 = opposite/hypotenuse

1/2 = x/6 ∵ sin 30 = 1/2

1/2 × 6 = x

6/2 = x

3/1 = x

Therefore, the length of x=3/1

4 0

2 years ago

The ________________ measures the spread of the middle 50% of the data set.

Scilla [17]

Answer:

C. interquartile range

Step-by-step explanation:

using this to show im right

"The interquartile range (IQR) is the difference between the upper (Q3) and lower (Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest."

Hope This Helped

6 0

1 year ago

What is the value of x?

Hoochie [10]

Answer:

xxxxxxxx

Step-by-step explanation:

x

3 0

2 years ago

Read 2 more answers

Please help I have 15 minutes!!!!!

OlgaM077 [116]

It is a function because each x value only has one y value

6 0

2 years ago

Read 2 more answers

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