Answer:
B the range, the x- and y-intercept
Step-by-step explanation:
the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).
but the range changes, as for the original function y could only have positive values - even for negative x.
the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.
the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.
the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.
the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.
the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)
Answer:
Coordinates: (0,0) ; (1,5) ; (2,10)
Step-by-step explanation:
x | y
0 0
1 5
2 10
Coordinates: (0,0) ; (1,5) ; (2,10)
This should be a straight line that is going diagonally with a positive slope.
Answer:
The inference that can be made using the dot plot is:
The range of round 1 is greater than the round 2 range.
Step-by-step explanation:
<u>Round 1:</u>
Score Frequency
1 0
2 2
3 3
4 2
5 1
Hence, the minimum score of Round 1 is: 2
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-2
=3
Similarly, <u>Round-2</u>
Score Frequency
1 0
2 0
3 0
4 4
5 4
Hence, the minimum score of Round 1 is: 4
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-4
=1
The scores of round 2 are higher than round-1.
Since round 2 have a higher frequency for higher scores as compared to round-1.
Hence, Range of round 1 is greater than the range of Round-2.
Histograms are useful when we have data which can be divided into several classes or groups. The histogram shows the trend of each class and the trend among the different classes. For example when we have about 50 different values ranging from 1 to 20, it will be a better approach to draw a histogram in this case by dividing the data into small ranges e.g 1 to 4, 5 to 9 and so on and counting the frequency for each class.
Dot plot is useful when we have a small number of individual values. In this case we can visualize how many times each individual value occurred in the data. This is useful when the number of values in the data is less.
In the given scenario, we have 12 values in total ranging from 1 to 5. So making a dot plot would be the best choice. A histogram would not be useful in this case.
Therefore, the correct answer is option D. Dot plot, because a small number of scores are reported individually
Answer:
x = 5
y = 1
Step-by-step explanation:
So move 2y from the second equation to the other side of the equals sign
2x +3y = 13
x = 3 + 2y
From here substitute X into the first equation
2(3 + 2y) + 3y = 13
Distribute
6 + 4y +3y = 13
Combine like terms and solve for y
7y = 7
y = 1
insert result into the original equation.
x-2(1) = 3
x - 2 = 3
x = 5
insert into other equation to check work
2(5) + 3(1) = 13
10 + 3 = 13 correct
