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san4es73 [151]
3 years ago
13

Let an integer be chosen at random from the integers 1 to 30 inclusive. Find the probability that the integer chosen is not even

ly divisible by 2.
Mathematics
2 answers:
WARRIOR [948]3 years ago
5 0

Answer:

probability that the integer not evenly divisible by 2 is  \frac{1}{2}.

Step-by-step explanation:

Given : An integer be chosen at random from the integers 1 to 30 inclusive.

To find  : Find the probability that the integer chosen is not evenly divisible by 2.

Solution : We have integers 1 to 30.

We need to find probability that the integer not evenly divisible by 2.

We know that odd numbers are not divisible by 2

Odd number 1 to 30 ( 1 , 3, 5, 7 ,9,11,13,15,17,19,21,23,25,27,29)

Total odd number 1 to 30 =  15 .

Probability =  \frac{favourble\ outcome}{total\ numver\ of\ outcome}.

Probability = \frac{15}{30}.

Probability = \frac{1}{2}.

Therefore,  probability that the integer not evenly divisible by 2 is  \frac{1}{2}.

Nataly [62]3 years ago
3 0
Exactly half of these integers are odd (1, 3, 5, ..., 29) so the probability of selecting an odd number is \dfrac12.
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A+b=6.1 ---(1)
a-b=1.6 ---(2)

(1)+(2); 2a=7.7
then a=3.85
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3 years ago
The bases of a trapezoid are 16.8 yards and 6.9 yards. what is the average of the two bases?
kow [346]
The average = 1/2 (sum of the two bases) = 1/2 ( 16.8 + 6.9) = 11.85 yards
3 0
3 years ago
Read 2 more answers
How many 3 digit numbers are possible when a) the leading digit cannot be zero and the number must be a multiple of 4?
guajiro [1.7K]

Step-by-step explanation:

I assume the digits can be repeated.

so, e.g. 555 is a valid number for this problem, right ?

that means we start with permutations with repetition :

n^r

n = the total number of items to pick from.

r = the number of items being picked per result.

we have 10 digits (0,1,2,3,4,5,6,7,8,9), and we pick 3 of them.

that gives us (with very little surprise, I hope)

10³ = 1000 different possible numbers from 000 to 999.

from these numbers we eliminate all with leading 0.

as we handled all digits the same way and with the same priority, there is the same amount of numbers for every digit in the leading position.

that means 1/10 of the total amount of numbers has a leading 0, or a leading 1, or a leading 2, ...

so, we need to subtract 1/10 × 1000 from 1000 :

1000 - 1000×1/10 = 1000 - 100 = 900

that would be the numbers 100 to 999.

and we have one more condition : the number must be a multiple of 4.

how many are there ?

well, that's the funny thing about numbers : from all numbers 1/2 of them are multiples of 2 (or divisible by 2), 1/3 of them are multiples of 3 (or divisible by 3), and ... you guessed it, 1/4 of them are multiples of 4 (or divisible by 4). and so on.

and so, 1/4 of our 900 numbers are multiples of 4 :

1/4 × 900 = 225

so, there are 225 possible 3-digit numbers that are multiples of 4 and do not start with a 0.

6 0
2 years ago
Need help on this question i dont get it ima be posting some more but i need help on this on first
Anuta_ua [19.1K]

Answer:

The answer is B.

Step-by-step explanation:

Its B because in the problem it says $12 per yard which means it would be 12y. The 40 Rupesh already had so just add it and it would be 12y+ 40. The key word "At least" results into the sign being greater than or equal to. So the final answer would be 12y+40\geq 148.

8 0
3 years ago
What value of "a" makes the system 3x-y=17 and ax+y=-7 have no solution?
Norma-Jean [14]

Answer:

y=-7-ax

3x-(-7-ax)=17

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3x+ax=10

x(3+a)=10

x=10,3+a=10 a=10-3 a=7

5 0
3 years ago
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