How many different integers between $100$ and $500$ are multiples of either $6,$ $8,$ or both?
1 answer:
We need to find the number of integers between 100 and 500 that can be divided by 6, 8, or both. Now, to do this, we must as to how many are divisible by 6 and how many are multiples of 8.
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101
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Answer:
The conditions necessary for the analysis are met.
The probability the newspaper’s sample will lead them to predict defeat is 0.7881
Step-by-step explanation:
We are given;
population proportion; μ = 52% = 0.52
Sample size;n = 400
The conditions are;
10% conditon: sample size is less than 10% of the population size
Success or failure condition; np = 400 x 0.52 = 208 and n(1 - p) = 400(1 - 0.52) = 192.
Both values are greater than 10
Randomization condition; we assume that the voters were randomly selected.
So the conditions are met.
Now, the standard deviation is gotten from;
σ = √((p(1 - p)/n)
where;
p is the population proportion
n is the sample size
σ is standard deviation
Thus;
σ = √((0.52(1 - 0.52)/400)
σ = √((0.52(0.48)/400)
σ = 0.025
Now to find the z-value, we'll use;
P(p^ > 0.5) = P(z > (x - μ)/σ)
Thus;
P(p^ > 0.5) = P(z > (0.5 - 0.52)/0.025)
This gives;
P(p^ > 0.5) = P(z > - 0.8)
This gives;
P(p^ > 0.5) = 1 - P(z < -0.8)
From the table attached we have a z value of 0.21186
Thus;
P(p^ > 0.5) = 1 - 0.21186 = 0.7881
Thus, the probability the newspaper’s sample will lead them to predict defeat is 0.7871
