Examine the graph.
The graph looks like one of square-root....let's see.
If we make a translation of (-3,-2), i.e. the beginning of the graph, to the origin, we will see that we have the following points on the graph,
(0,0), (1,1), (4,2), (5,2.2), (8,2.8)
all of which correspond to the sqrt(x) function.
Thus if we take the parent graph as sqrt(x), and translate to the current position by a translation of (-3,-2), we get the current graph.
To do the translation, we recall that a translation of a parent function f(x) by (h,k) can be effected by
g(x) ->f(x-h)+k
Setting h=-3, k=-2, and f(x)=sqrt(x), we get
current graph g(x)=f(x-(-3)+(-2)=f(x+3)-2 = sqrt(x+3)-2
Answer:
☑ Answer is 6
Step-by-step explanation:
» But this is a ratio, so it can be expressed in terms of fractions:
» Divide both the denominator and numerator by 9:
Answer:
-3°F
Step-by-step explanation:
6 - 9 = -3°F
Answer:
<em>5.5</em>
Step-by-step explanation:
Given the set of data
5, 4, 2, 1, 1, 2, 10, 2, 3, 5.
The average of the least and the greatest value is known as the midrange
The formula for calculating the midrange is expressed as shown:
Midrange = (Greatest value + Least value)/2
Given
Greatest value = 10
Least value = 1
Midrange = 10+1/2
Midrange = 11/2
Midrange = 5.5
<em>Hence the midrange of the data is 5.5</em>
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year
