Suppose that the <u>function</u> f(x) is the <u>parrent</u> function and the<u> graph</u> of the function g(x)=f(x)-a can be obtained from the graph of the parrent function f(x) by <u>shifting down</u> a units.
Rewrite the expression for the function 
in the following way:
This shows that the shift down is made by 17 units.
Answer: 17 units
Answer:
Type I error occurs when the null hypothesis, H0, is rejected, although it is true.
Here the null hypothesis, H0 is:
H0: Setting weekly scheduled online interactions will boost the well being of people who are living on their own during the stay at home order.
a) A Type I error would be committed if the researchers conclude that setting weekly scheduled online interactions will not boost the well being of people who are living on their own during the stay at home order, but in reality it will
b) Two factors affecting type I error:
1) When the sample size, n, is too large it increases the chances of a type I error. Thus, a sample size should be small to decrease type I error.
2)A smaller level of significance should be used to decrease type I error. When a larger level of significance is used it increases type I error.
If there’s a total of 210 fruit trees, and each tree produces 590 pounds a year, you multiply both those numbers and get 123,900 pounds a year.
Option C is correct option.
Step-by-step explanation:
We need to find the solution of 
The given inequality becomes undefined when x = 1 because the denomiator is: 1-x and if x = 1 then it becomes 1-1 = 0 and anything divided by zero is undefined.
So, all values other than 1 are included in the solution of the given inequality.
So, solution is: x<1 or x>1 but x≠1
So, Option C is correct because all numbers are included in number line except 1. An unfilled circle on 1 shows that it is not included in the solution. Rest of numbers are included.
Option C is correct option.
Keywords: Solving inequalities
Learn more about Solving inequalities at:
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Answer:
B. 2/3
Step-by-step explanation:
To solve this we have to take into account this axioms:
- The total probability is always equal to 1.
- The probability of a randomly selected point being inside the circle is equal to one minus the probability of being outside the circle.
Then, if the probabilities are proportional to the area, we have 1/3 probability of selecting a point inside a circle and (1-1/3)=2/3 probability of selecting a point that is outside the circle.
Then, the probabilty that a random selected point inside the square (the total probability space) and outside the circle is 2/3.
