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kotykmax [81]
3 years ago
10

Which one is greater -1.7 or -1

Mathematics
1 answer:
myrzilka [38]3 years ago
4 0

Answer: -1

Step-by-step explanation: -1 is to the right of 1.7 and if anything is to the right is greater (when it is dealing with negative numbers)

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100+29=129

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What is the probability of spinning a number greater than 7 or an odd number? (1 Point)
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Find the explicit formula that produces the given sequence. <br> 3/2, 3/4, 3/8, 3/16,...
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\frac{3}{2};\ \frac{3}{4};\ \frac{3}{8};\ \frac{3}{16};\ ...\\\\a_1=\frac{3}{2}\\\\a_2=\frac{3}{4}=\frac{3}{2}\cdot\frac{1}{2}\\\\a_3=\frac{3}{8}=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{2}\cdot\left(\frac{1}{2}\right)^2\\\\a_4=\frac{3}{16}=\frac{3}{8}\cdot\frac{1}{2}=\frac{3}{2}\cdot\left(\frac{1}{2}\right)^3\\\vdots
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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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