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Ad libitum [116K]
3 years ago
5

PLEASE HELP ME BY ANSWERING AT LEAST 2 QUESTIONS MY WORK IS DUE IN LESS THAN AN HOUR!

Mathematics
1 answer:
levacccp [35]3 years ago
8 0

nononononon nonononononno

Step-by-step explanation:

nonononononomonjknonkkom

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If two events are complementary, then we know that: Multiple Choice the sum of their probabilities is one. the joint probability
notka56 [123]

Answer:

The joint probability of the two events is one.

Step-by-step explanation:

Complementary events:

If two events are complimentary, these three following things are true:

They are dependent.

The intersection of them is zero.

The joint probability of the two events is one.

The last one is the correct choice.

3 0
3 years ago
I WILL GIVE BRAINLEST FOR THE CORRECT ANSWER!
pentagon [3]

Answer:

A) -4

B) 4 units apart

C) no values of x

Step-by-step explanation:

1) slope of m: (3-2)/(10-8) = ½

m(x) = ½x + c

2 = ½(8) + c

c = -2

m(x) = ½x - 2

m(16) = 6

h(4) = (4-2)²/2 = 4/2 = 2

h(4)-m(16) = 2-6 = -4

B) y-intercept of:

*h(x) = ½(0-2)² = 2

*m(x) = -2

from the equation found before

2 - (-2) = 4 units apart

C) m(x) = (x-4)/2

m(x) > h(x)

(x-4)/2 > ½(x-2)²

x-4 > x²-4x+4

x²-5x+8 < 0

(x - 2.5)² + 1.75 < 0

Never negative

3 0
3 years ago
What is the greatest common factors of 12 and 27
aalyn [17]
12 is divisible by 1,2,3,4,6,12 and 27 is by 1,3,9,27 so the GCF is 3
3 0
3 years ago
Write an equation if you have the points (3,6) and (3,-4)
Oduvanchick [21]

Answer: y=6x-12

Step-by-step explanation:

5 0
3 years ago
The base of an auditorium is in the form of an eclipse 200 feet long and 100 feet wide a pin drop near one focus can clearly be
dedylja [7]

Answer:

Let the coordinate of focus be (\pm c , 0)

As per the statement: The base of an auditorium is in the form of an eclipse 200 feet long and 100 feet wide.

⇒Length of Major axis=base of an auditorium = 200 feet and Length of a minor axis=wide of a auditorium = 100 ft

Semi-major axis (a) = 100 ft and

semi-minor axis(b) = 50 ft

Then, by an equation:

c^2 = a^2-b^2

Solve for c:

Substitute the given values we have;

c^2=(100)^2-(50)^2

Simplify:

c^2 = 7500

or

c=\sqrt{7500} = 86.6025404 ft

Distance between the foci is,  2c = 2 \cdot 86.6025404 = 173.205081

Therefore, the distance between the foci to the nearest 10th of a foot is, 173.2 ft

6 0
3 years ago
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