All categories

3 years ago

Consider parallelogram ABCD. Choose all of the statements which must be true

Mathematics

2 answers:

Crazy boy [7]3 years ago

5 0

The answers are 1, 2, and 4. Look at the parallelogram's angles to be sure.

stiv31 [10]3 years ago

3 0

Answer:

$AE=EC$

$\angle ADB \cong \angle CBD$

$m\angle ADC + m\angle DCB = 180\°$

Step-by-step explanation:

In the given parallelogram, we know that

$AD \parallel BC\\AB \parallel DC$

From this parallels and having segment $DB$ as a transversal, we have

$\angle ADB \cong \angle CBD$ by alternate interior angles.

Then, we have

$AE=EC$ by givens.

Finally, we have

$m\angle ADC + m\angle DCB = 180\°$ , by consecutive interior angles, which are supplementary by definition.

Therefore, the right answers are

$AE=EC$

$\angle ADB \cong \angle CBD$

$m\angle ADC + m\angle DCB = 180\°$

You might be interested in

What is the answer to this?

konstantin123 [22]

Answer:

$\huge\boxed{-6}$

Step-by-step explanation:

Since we know the values of a, b, and c, we can substitute them into our expression and get the value.

$-2b + | -3c + a |\\\\-2(6) + |-3(-3) + -3 |$

A negative multiplied by a negative gets us a positive.

$-2(6) + |9 + -3 |$

Adding a negative is the same as subtracting a positive.

$-2(6) + |9 -3 |\\-2(6) + 6$

$-2 \cdot 6 = -12\\\\-12+6\\\\-6$

Hope this helped!

3 0

3 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