AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?

Is this right I am positive but I need to pass this test

FrozenT [24]

Answer:

It's not a polygon

Step-by-step explanation:

Polygons are always 2D. This shape is 3D. Everything else is fine.

Please mark as Brainliest! :)

Have a nice day.

3 0

3 years ago

Find the 76th term of the arithmetic sequence 17 19 21

OLga [1]

Answer:

167

Step-by-step explanation:

4 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June

Shtirlitz [24]

Answer:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2 +10$

Alternative hypothesis: $\mu_1 >\mu_2 +10$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 10$

Alternative hypothesis: $\mu_1 -\mu_2>10$

Our notation on this case :

$n_1 =129$ represent the sample size for group AISI 1078

$n_2 =129$ represent the sample size for group AISI 1064

$\bar X_1 =123.6$ represent the sample mean for the group AISI 1078

$\bar X_2 =107.6$ represent the sample mean for the group AISI 1064

$s_1=2.0$ represent the sample standard deviation for group 1 AISI 1078

$s_2=1.3$ represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

Now we can calculate the degrees of freedom given by:

$df=129+129-2=256$

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

7 0

3 years ago

Why is it helpful to estimate the answer before adding or subtracting decimals?

AlekseyPX

So you can get a basic idea of what your actual answer will be in range of.

Example:
10.5+5.5=__

I can safely assume with estimating that it'll be 15.
Now when I actually find the correct answer which is 16, its a safe bet knowing that your estimate is close to your actual answer.

If you make an uneducated guess like saying 10.5+5.5 was 30 then your answer was 16 its makes you wonder how you got an estimate of almost 2x of 16.

Estimates help, so you know what range your real answer would be in.

Hope this helps

4 0

3 years ago

The perimeter of a △ABC equals 12 in. The midpoints of the sides M, N and K are connected consecutively. Find the perimeter of △

Feliz [49]

Answer:

6 inches

Step-by-step explanation:

Triangle midline theorem: In a triangle, the midline joining the midpoints of two sides is parallel to the third side and half as long.

Points M, N and K are midpoints of the sides AB, BC and AC. Acording to the Triangle midline theorem,

$MN=\dfrac{1}{2}AC,\\ \\NK=\dfrac{1}{2}AB,\\ \\MK=\dfrac{1}{2}BC.$

The perimeter of the triangle MNK is

$P_{MNK}=MN+NK+MK=\dfrac{1}{2}AC+\dfrac{1}{2}AB+\dfrac{1}{2}BC=\dfrac{1}{2}(AC+AB+BC)=\dfrac{1}{2}P_{ABC}=\dfrac{1}{2}\cdot 12=6\ in.$

5 0

3 years ago

AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?​

AB is perpendicular to CD. If the slope of AB is 2/5, what is the slope of CD?