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Brilliant_brown [7]
3 years ago
11

If a bracelet takes 5/6 foot of string how much string will it take to make 12 bracelets

Mathematics
1 answer:
const2013 [10]3 years ago
3 0
5/6 x 12


60 / 6  = 10

It takes 10 feet of string to make 12 bracelets.

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153 is 0.9% of what number tell which equivalent ratios are used to find a solution
Klio2033 [76]

Answer: Answer: 153 is 0.9% of 17000.

(brainly please)

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May I please have help from anyone thank you so much !
FrozenT [24]

Answer: Choice D

======================================================

Explanation:

The formula for the surface area of a cylinder is

SA = 2\pi*r^2 + 2\pi*r*h

We then replace r with 10 and h with 25. Keep in mind that the diameter is 20 mm, so the radius is half that at 10 mm.

So that's how we get to

SA = 2\pi*10^2 + 2\pi*10*25

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2 years ago
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Find the probability of the following events , when a dice is thrown once:
Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(c) Let A be an event ''a number greater than 4''.

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

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3 years ago
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Which description best fits the graph?
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Answer:

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2 years ago
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The larger of two number is 15 more than three times the smaller number if the sum of the two numbers is 63 find The numbers
Aleks [24]

The two numbers are <u>12 and 51.</u>

<h3>EXPLANATION</h3>

To solve this, I did 63-15, to get 48.

I then divided this by 4, to get the first number, which is <u>12.</u>

To find the second answer, I multiplied 12 by 3, which is 36, and then added the 15 back on, to get <u>51.</u>

<u>51 + 12 = </u><em><u>63</u></em>

3 0
3 years ago
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