3 = -(-y + 6)
First, simplify brackets. / Your problem should look like: 3 = y - 6
Second, add 6 to both sides. / Your problem should look like: 3 + 6 = y
Third, simplify 3 + 6 to 9. / Your problem should look like: 9 = y
Fourth, switch sides. / Your problem should look like: y = 9
The answer is D) 9.
Start with 180.
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>
<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>
<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>
<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>
<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>
<span>Once you end up with a quotient of "1" you're done. </span>
<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
Let’s give these two numbers variables
Let ‘a’ be the larger number
Let ‘b’ be the smaller number
Now from the question we know:
a + b = 30 or a = 30 - b
2a - 3b = 5
Now, let’s plug the first equation into the second to find ‘b’:
2(30 - b) - 3b = 5
60 - 2b - 3b = 5
60 - 5b = 5
55 = 5b
b = 11
Now we solve for ‘a’:
a = 30 - b
a = 30 - 11
a = 19
Now the question asks for the positive difference between the two numbers, so:
a - b = ?
19 - 11 = 8
Hope this helps!
Answer:
The mean is 95 and the standard deviation is 2
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean 
and standard deviation 
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
Population:
Mean 95, Standard deviation 12
Samples of size 36:
By the Central Limit Theorem,
Mean 95
Standard deviation 
