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Nimfa-mama [501]
2 years ago
5

De cuántas formas pueden mezclarse los siete colores del arco iris tomándolos de tres en tres?

Mathematics
1 answer:
Tanzania [10]2 years ago
5 0

Answer:

hi

Step-by-step explanation:

1+1=2

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If anyone can answer this, I will give you hearts!
kotegsom [21]

Answer:

  see attached

Step-by-step explanation:

Rotation 270° counterclockwise is equivalent to rotation 90° clockwise. The transformation of coordinates is ...

  (x, y) ⇒ (y, -x) . . . . . . . rotation 270° CCW

This means the points are moved to ...

  A(-2, 1) ⇒ A'(1, 2)

  B( 1, 2) ⇒ B'(2, -1)

  C(-2, 4) ⇒ C'(4, 2)

The rotated triangle is shown in the attachment. You may notice that A' and B are the same point.

6 0
2 years ago
What is the value of 780/60
a_sh-v [17]

Answer: 13

Step-by-step explanation: If you add 13 to itself 60 times, you will get 780.

5 0
2 years ago
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Which statement is true about this function h(x) = 2^x-5
Marysya12 [62]

Answer:

\frac{1}{32} x

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Which expression can be used to find the area of the rectangle shown on the coordinate plane?
Naily [24]
Answer:

8 * 7

Explanation:

When you count the squares along the area, you will find that the base is 8 units, and the height is 7 units. The area of a square is equal to base * height, so the correct answer is 8 * 7.
7 0
3 years ago
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From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x ≥ 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
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