suppose X and Y are independent random variables, both with normal distributions. If X has a mean of 45 with a standard deviatio
1 answer:
Answer:
0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question:
What is the probability that a randomly generated value of X is greater than a randomly generated value of Y
This means that the subtraction of X by Y has to be positive.
When we subtract two normal variables, the mean is the subtraction of their means, and the standard deviation is the square root of the sum of their variances. So
We want to find P(X > 0), that is, 1 subtracted by the pvalue of Z when X = 0. So
has a pvalue of 0.0228
1 - 0.0228 = 0.9772
0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y
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Answer:
Step-by-step explanation:
So lets go over what we know:
There are 3 large pizzas
Each pizza has a addition cost of 2 dollars
The total cost is 42 dollars
We need to find the cost per pizza.
Now lets piece together our equation.
We know that the pizzas equal 42 dollars so we can immediately put 42 on the opposite side of the equal side:
We also know that each of the 3 pizzas cost 2 dollars. This would go on the left side of the equation:
Now, we know x is the amount of pizzas. There are 3 pizzas, or in other words, 3x. We can add this to the left side of the equation, because the 3 pizzas plus the 6 dollars equals the total of 42 dollars:
There are no answers above that look like this, however. This is because we have to factor the left side of the equation.
A factor of both 3 and 6 is 3. So we can factor out 3 and we get:
This looks like the first answer!
Hope this helps! :3
Note that powers of 2 can be written in binary as
and so on. Observe that
digits are required to represent the 
-th power of 2 in binary.
Also observe that
so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,
That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of
. We only need the integer part, 
, then we add 1.
Now,
, and 999 falls between these consecutive powers of 2. That means
which means 999 requires
binary digits.
Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write
? Or do you pad this number with 0s to get 10 digits, i.e. 
? In the latter case, the answer is obvious; 
total binary digits are needed.
In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
and so on. Observe that
Also observe that
so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,
That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of
Now,
which means 999 requires
Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write
In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
Answer:
A, D , and E
Step-by-step explanation:
We have that:
is the image of CD after a dilation of scale factor n.
We use the relation between the image length and object length:
Option A
If n=3/2, then
This is true.
Option B
If n=4, then
This is false.
Option C
If n=8, then
This too is false.
Option D
If n=2, then
This is true
Option E
If n=3/4, then
This is also true.
