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Nikitich [7]
3 years ago
14

A child sitting on the edge of a merry go round travels 22m in one full turn, What is the diameter of the merry-go-round?​

Mathematics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

Diamter=7 meter

Step-by-step explanation:

One full turn means we have 360° of theta which equals 2pi in radinas

And arc length is 22 so using the equation theta= L/r = 2pi=22/r so r= 3.50 and diameter= 2r =7 m

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)) Matt's bill for breakfast at a restaurant was $91. He left an 18% tip. What was the total
gladu [14]

Answer:

$107.38

Step-by-step explanation:

$91 (the price) multiplied by the percentage 18% (91 x 18 = 1630) (1638/100=16.38 [the tip})

$91 (th original price) + 15.38 (the tip) = $108.38 (final price)

4 0
3 years ago
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7) Is -2 a zero (or root) of f(x) = 7x3 + x2 - 9x + 5?
Masja [62]

Answer:

Step-by-step explanation:

i do not know

6 0
3 years ago
3.0 plzzzzzz help thank you
hoa [83]

Answer:

80%

Step-by-step explanation:

100/5*4

8 0
3 years ago
How do you determine the value of isosceles triangles?
amid [387]

There's really no such thing as the value of a triangle. 

Every triangle has three sides, three angles, a base, a height, and an area,
and there could be problems that ask us to find any one of those. 

Whatever we need to find, the process is always the same: 

-- Take the information that's given.

-- Gather up everything you can remember that talks about a relationship
between what you're given and what you need to find.

-- Use them together to find the missing value.


6 0
3 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
3 years ago
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