Since the 90% is 63. divide it by 9. then you would get value of 7. then multiple it by 10.
which is equal to 70. so 100% is 70.
<span>f(x) = 4x + 3
g(x) = -2x + 5
(f o g) (5) means the function of g when g is a function of 5
(the function of the function concept)
so we need to get g(5) firstly
</span>
<span>g(x) = -2x + 5
g(5)= -2(5) + 5 = -5
After that, f(g(5)) = f(-5) = 4(-5) + 3 = -20 + 3 = -17
I hope that helps
</span>
Remember that the difference between census and a sample survey is that in a census you take all the people of population in consideration in the study, and in the sample survey you take a part of that population in consideration to survey.
SO here if you check the problem it says that "in a recent poll, Pew reserchers found that 47% OF AMERICAN ADULT RESPONDENTS"
that is the population of the study that you want to focus, but there are a billion of american adult in US right? so its impossible that this company Pow Research made a census if you consider that a billion of people in US is almost infinite number for a company with a finite number of employees.
Also remember that a company has a budget to consider every month or every year, so to make a census or a sample survey it takes time and money, just to survey 10 people in a streeth of US cost money so if you want to reduce cost you might need to do a sample survey instead of a census.
Answer:
$10.80
Step-by-step explanation:
Given:
Total number of CDs = 6
Cost price (including tax) = $71.40
Each CD had a tax of $1.10
To find the price of each CD before tax Let's first find the price of each CD after tax.
Price after tax =
Price of each CD after tax = $11.90
Since each CD had a tax of $1.10, price of each CD before tax would be : $11.90 - $1.10 = $10.80
Therefore price of each CD before tax is $10.80
Answer:
Check the explanation
Step-by-step explanation:
Hypotheses are:
Following is the output of t test:
Hypothesis Test: Independent Groups (t-test, pooled variance)
X1 X2
12 15.5 mean
3.6591 3.6839 std. dev.
10 15 n
23 df
-3.5000 difference (X1 - X2)
13.4999 pooled variance
3.6742 pooled std. dev.
1.5000 standard error of difference
0 hypothesized difference
-2.33 t
.0287 p-value (two-tailed)
The p-value is:
p-value = 0.0287
Since p-value is less than 0.05 so we reject the null hypothesis. That is we can conclude that the exercise program have a significant effect.
(b)
The cohen's d is