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const2013 [10]
3 years ago
15

Please answer this question now

Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

AB = 72°

Step-by-step explanation:

The inscribed angle ADC is half the measure of its intercepted arc, thus

56° = \frac{1}{2} ( m ABC ) ← multiply both sides by 2

112° = ABC

ABC = AB + BC = AB + 40, so

AB + 40 = 112 ( subtract 40 from both sides )

AB = 72°

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Answer:

2940

Step-by-step explanation:

You are subtracting a negative number. "Minus a negative is plus."

525 - (-2415) = 525 + 2415 = 2940

6 0
2 years ago
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The cube less the square of a number is twice the number. If x is a positive integer, what is its value
olga55 [171]

Answer:

2

Step-by-step explanation:

let 'x' = number

x³ - x² = 2x

x³- x² - 2x = 0

x(x² - x - 2) = 0

x(x - 2)(x + 1) = 0

x = -1, 0, 2

the only positive solution is the number 2

3 0
2 years ago
Use the spinner below to find the probability of getting the following number after 1 spin. P(multiple of 3) = (Round to 4 decim
Talja [164]

The value of the probability P(multiple of 3) is 0.3333

<h3>How to determine the probability?</h3>

The spinner that completes the question is added as an attachment

From the attached spinner, we have:

  • Total section = 12
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The probability is then calculated as:

P(multiple of 3) = Multiples of 3/Total

This gives

P(multiple of 3) = 4/12

Evaluate

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Hence, the value of the probability is 0.3333

Read more about probability at:

brainly.com/question/25870256

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3 0
2 years ago
Taylor took 560 photographs during summer vacation. She placed 12 photos on each page of her scrapbook, but she had fewer than 1
Hoochie [10]
Taylor placed 8 photos on the last page of her scrapbook.
4 0
3 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
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