Which is the graph of f(x) = (x – 1)(x + 4)?
2 answers:
ANSWER
See attachment.
EXPLANATION
The given function is
f(x) = (x – 1)(x + 4).
This parabola will open upwards because the leading coefficient is positive.
The x-intercepts can be found by equating the function to zero.
By the zero product property;
This implies that,
The graph that touches the x-axis at -4 and 1, and opens upwards is the last graph.
The correct choice is D.
Answer:
The fourth graph (last graph)
Step-by-step explanation:
Remember that the zeros of a function are the x-intercepts of the graph. To find the zeros we just need to set the function equal to zero and solve for x:
Now we know that the graph or our function intersects the x-axis at x = 1 and x = -4.
Since both x values inside the parenthesis are positive, our parabola is opening upwards.
The only graph opening upwards whose x-intercepts are x = 1 and x = -4 is the fourth one.
We can conclude that the graph of is the fourth one.
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Answer:
B) 4.07
Step-by-step explanation:
First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:
Regular BelowRegular Premium SuperPremium
39.31 36.69 38.99 40.04
39.87 40.00 40.02 39.89
39.87 41.01 39.99 39.93
X1⁻=39.68 X2⁻= 39.23 X3⁻= 39.66 X4⁻= 39.95
Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63
Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):
SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12
SSW = 11.12
Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom
SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974
SSB = 0.80
For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom
Now we can establish the hypothesis for the test:
H0: μ1 = μ2 = μ3 = μ4
The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:
H1: the grade of gasoline does makes a difference
We will use the F statistic to test the hypothesis, which is calculated like follows:
F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19
We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:
Fc = 4.07
