Answer:
2 should be distributed as 2y + 8; y = 4
Step-by-step explanation:
2(y + 4) = 4y
Distribute
2y + 8 = 4y
Step 2 is incorrect because the added instead of multiplied
Continuing on to correct the problem
Subtract 2y from each side
2y+8-2y = 4y-2y
8 = 2y
Divide by 2
8/2 = 2y/2
4 =y
1,54 3,8 6,9
multiply the pairs of numbers
Answer:
We have that the sum of two numbers is 9
this can be written as:
x + y = 9
where x is the larger number.
Now we want to write:
"the difference between one more than the larger number and twice the smaller number"
First, remember that the difference between A and B is:
A - B
Then "the difference between one more than the larger number and twice the smaller number"
is:
"one more than the larger number" = ( x + 1)
"twice the smaller number" = 2*y
the difference between these is:
(x + 1) - 2*y
Now we can simplify:
We know that:
x + y = 9
then:
y = 9 - x
replacing that in the equation:
(x + 1) - 2*y
we would get:
x + 1 - 2*(9 - x)
x + 1 -18 + 2x
(x + 2x) + (1 - 18)
3x - 17
This means that we can write:
"the difference between one more than the larger number and twice the smaller number"
as: 3x - 17
The answer is b bc it’s all real numbers since it won’t stop
Answer:
A. The larger the sample size the better.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes 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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
In this question:
We have to look at the standard error, which is:
This means that an increase in the sample size reduces the standard error, and thus, the larger the sample size the better, and the correct answer is given by option a.