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Aleksandr-060686 [28]
2 years ago
15

In the document can someone please help me understand it

Mathematics
1 answer:
V125BC [204]2 years ago
7 0

Answer: I see two people typing, so...

Step-by-step explanation:

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A bus travels with a constant speed of 60 miles per hour. How long will it take to travel 75 miles?
sukhopar [10]

Answer:

a total of 4,500

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2 years ago
Select all that apply. Several pennies are tossed in the air. Twelve land heads up and fifteen land tails up. Which of the follo
jeka94

Answer:

12 heads to 15 tails

Step-by-step explanation:

Heads : Tails

12 : 15

4 : 5

4 0
3 years ago
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Help due tonight.. *no links please*
elena-s [515]

Answer:

measure of angle 3= 109

angle 2=71

angle 1=109

Step-by-step explanation:

a straight line =180

7 0
3 years ago
12% of 72 is what number ?
valina [46]
The best thing to do here is to find 1%, then multiply this by 12.
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Therefore, 12% of 72 is 8.64
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6 0
3 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

7 0
3 years ago
Read 2 more answers
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