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2 years ago

Which of the following statements help prove that calypso rd and capoeira rd are parallel?

Mathematics

2 answers:

eduard2 years ago

8 0

Answer: The answers are given below.

Step-by-step explanation: We are given two straight lines calypso rd and capoeira rd. Both are cut with a transversal Lambda Ave. We are select all the correct conditions from the given options that will help to prove that lines calypso rd and capoeira rd are parallel to each other.

Condition (1) is

$m\angle 2=81^\circ,~~m\angle 3=89^\circ.$

Since ∠2 and ∠3 are interior angles on the same side of the transversal and their sum is 81° + 89° = 170° ≠ 180°, so this condition will not help.

Condition (2) is

$m\angle 1=90^\circ,~~m\angle 5=90^\circ.$

Since ∠1 and ∠3 are vertically opposite angles and ∠3 and ∠5 are corresponding angles, so we have m∠1 = m∠3 = m∠5, so this condition will help.

Condition (3) is

$m\angle 4=85^\circ,~~m\angle 5=85^\circ.$

Since the measurement of ∠4 and ∠5 do not have anything related to prove the parallelism of the given lines, so this condition will not help.

Condition (4) is

$m\angle 2+m\angle 6=180^\circ.$

Since ∠2 and ∠6 are corresponding angles, so their measurement shoul be equal in order to prove the lines parallel. vAs our given condition does not imply this, so this condition will also not help.

Condition (5) is

$m\angle 2+m\angle 3=180^\circ.$

Since ∠2 and ∠3 are interior angles on the same side of the transversal, and their sum is given to be 180°, so this condition implies that the two lines are parallel. Hence, this condition will help.

Condition (6) is

$m\angle 3+m\angle 4+m\angle 5+m\angle 6=360^\circ.$

Since this is the general condition that the measurement of the whole angle is 360°, so this condition will not help.

Condition (6) is

$m\angle 1=m\angle 3=m\angle 5.$

Since ∠1 and ∠3 are corresponding angles and ∠3 and ∠5 are alternate interior angles, so their equality will definitely help in proving the lines to be parallel.

Thus, the correct conditions are

$m\angle 1=90^\circ,~~m\angle 5=89^\circ.$

$m\angle 2+m\angle 3=180^\circ.$

$m\angle 1=m\angle 3=m\angle 5.$

nydimaria [60]2 years ago

3 0

Did anyone reply? plzzzz let me know

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a ball is thrown with a slingshot at a velocity of 110ft/sec at an angle of 20 degrees above the ground from a height of 4.5 ft.

satela [25.4K]

Answer:

$t=2.47\ s$

Step-by-step explanation:

The equation that models the height of the ball in feet as a function of time is

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Where $h_0$ is the initial height, $s_0$ is the initial velocity and t is the time in seconds.

We know that the initial height is:

$h_0 = 4.5\ ft$

The initial speed is:

$s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s$

So the equation is:

$h (t) = 4.5 + 37.62t -16t ^ 2$

The ball hits the ground when when $h(t) = 0$

$4.5 + 37.62t -16t ^ 2 = 0$

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

$at^2 +bt + c$

The quadratic formula is:

$t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}$

In this case

$a= -16\\\\b=37.62\\\\c=4.5$

Therefore

$t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}$

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2 years ago

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: