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3 years ago

Which of the following are the zeros of the function y= 3(x-a)(x+b) in terms of a and b?

Mathematics

1 answer:

frosja888 [35]3 years ago

7 0

The correct answer is option 2.

3 = 0

x - a = 0

x = a

x + b = 0

x = -b

We do not multiply any of these values by 3 because in a way, 3 is acting as another "zero" of the equation. Though, as shown above, it does not equal 0, it still does not apply to any of the other equations solved above.

Hope this helps!! :)

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Name the line and plane shown in the diagram. A.AB and plane ABD B.BA and plane DC C.AC and plane ABD D.AB and plane DA

elena-14-01-66 [18.8K]

Answer: The line is AB and the plane is ABD, the first option is the correct one.

Step-by-step explanation:

Ok, first some definitions.

A line is any line that crosses two colinear points. Particularly, you can see in the graph that the line crosses through A and B, so the line is AB.

A plane needs 3 non-colinear points (if the points where colinear, then the points may define a line). Other definition of plane is "a line and a point that is not in the line"

So, if our line is AB, then the possible planes are:

ABC and ABD.

then the correct option is:

Line AB and plane ABD, so the correct option is the first one.

6 0

2 years ago

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline

yawa3891 [41]

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