Answer:
x+x+30 is the required expression.
Step-by-step explanation:
You watch x minutes of television on Monday
An since, the same amount on Wednesday that means x minutes on Wednesday
And 30 minutes on Friday
Since, we have to tell the expression that represents the total amount in the given situation that means
x+x+30 is the required expression.
Step-by-step answer:
The domain of log functions (any legitimate base) requires that the argument evaluates to a positive real number.
For example, the domain of log(4x) will remain positive when x>0.
The domain of log_4(x+3) requires that x+3 >0, i.e. x>-3.
Finally, the domain of log_2(x-3) is such that x-3>0, or x>3.
Answer:
Option A is right
Step-by-step explanation:
Given that approximately 52% of all recent births were boys. In a simple random sample of 100 recent births, 49 were boys and 51 were girls. The most likely explanation for the difference between the observed results and the expected results in this case is
A) variability due to sampling
-- True because there is a slight difference whichmay be due to sampling fluctuations.
B) bias
False because given that 100 random births selected
C) nonsampling error
False. There is no chance for systematic error here.
d) Confounding: There is no confounding variable present inchild birth since each is independent of the other
e) a sampling frame that is incomplete
False because the sampling is done correctly.
So, okay
$19.500 is 100%
wee need to find 1.5%
so it gonna be 19.500/100=195$ so it is 1%
half of procent will be 97.5
then 1.5% gonna be 195+97.5=292.5
for a year will be 19.792.5$
enjoy it
Answer:
All the three statements presented are indeed valid and satisfy the De Morgan's theorem.
Step-by-step explanation:
The truth tables are presented in the attached images.
a) The statement (X + Y + Z)' = (X'Y'Z') has its proof in the first image attached.
b) The statement (XYZ)' = (X' + Y' + Z') has its proof in the second image attached.
c) The statement (X + YZ) = (X + Y)(X + Z) has its proof in the third image attached.
