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damaskus [11]
3 years ago
14

NEED THE ANSWER ASAP PLEASE

Mathematics
1 answer:
Goryan [66]3 years ago
7 0

Answer:

x=6

Step-by-step explanation:

use trig identities

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Liz buys 6 flowerpots. there are 8 flowers in each pot. how many flowers did Liz buy?
vivado [14]

Answer:

48

Step-by-step explanation:

6 flowerpots

8 flowers per pot

8 x 6 = 48 flowers

4 0
3 years ago
Millie works 8 hours on Saturday and earns £6.50 an hour. Her bus fare to work costs £2.75 and she also spends £4.25 on her lunc
Nimfa-mama [501]
The answer is 7.50 ......
6 0
3 years ago
What is the answer to this question?
Klio2033 [76]
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7 0
3 years ago
Read 2 more answers
Find the number of primes less than 190 using the principle of inclusion-exclusion.
Troyanec [42]

The number of primes less than 190 using the principle of inclusion-exclusion are 42

<h3>Principle of inclusion- exclusion</h3>

The principle of inclusion-exclusion is known as a counting technique that computes the number of elements satisfying at least one of several properties and guaranteeing that the numbers are not counted twice.

Prime numbers are numbers only divisible by 1 and itself.

Prime numbers less than 190 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181

They are 42 in number

Therefore, the number of primes less than 190 using the principle of inclusion-exclusion are 42

Learn more about prime numbers here:

brainly.com/question/874965

#SPJ11

8 0
1 year ago
What is the probability of getting either a sum of 5 or at least one 5 in the roll of a pair of dice?
maria [59]
In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.

To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9

To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum).  The probability is 11/36

Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.

P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9
4 0
3 years ago
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