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Liula [17]
3 years ago
8

Please help me, I've been stuck on this problem forever and I need help! Please don't answer with something random to get points

. Answer the question please. I will mark Branliest.
Complete the two-column proof
Given: ∠1 is complementary to ∠2
→BD bisects ∠ADC
Prove: ∠1 is complementary to ∠3

Statement Reason
1. ∠1 is complementary to ∠2 1. ?

2. ? 2. Definition of complementary

3. ? 3. Given

4. ∠2 ≅ ∠3 4. ?

5. m∠2 = m∠3 5. ?

6. ? 6. Substitution property of equality

7. ? 7. ?

Mathematics
1 answer:
vodomira [7]3 years ago
8 0
1. given
2. measure angle 1 plus measure angle 2 equals 90 degrees
3. B D bisects angle A D C
4. definition of angle bisector
5. definition of congruence
6. measure angle 1 is equal to measure angle 3
7. angle 1 is congruent to angle 3
7. definition of congruence
hope that helps
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Are similar polygons congruent?
Vilka [71]

Polygons are similar when they have the same number of sides,
their corresponding angles are equal, and their corresponding
sides are all in the same ratio.

When that ratio is ' 1 ', they are also congruent.

That can happen sometimes, but it doesn't always have to happen.

5 0
3 years ago
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Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog
lara31 [8.8K]

Answer:

SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048

If we replace the values obtained we got:

0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257

0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

X=35 number of people that agree.

n=100 random sample taken

\hat p=\frac{35}{100}=0.35 estimated proportion of people that agree

p true population proportion of peopl that agree

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".  

The population proportion have the following distribution

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

Solution to the problem

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:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The standard error is given by:

SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048

If we replace the values obtained we got:

0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257

0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443

The 95% confidence interval would be given by (0.257;0.443)

6 0
3 years ago
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Simplify expression 2/3[3f+12]+f
neonofarm [45]

Answer:

= 3f + 8

Step-by-step explanation:

Given that:

= 2/3[3f+12]+f

By simplifying:

2/3 will be multiplied inside the bracket as follows:

= 2/3*3f + 2/3*12 + f

By cancelling the terms with each other we get:

= 2f + 24/3 + f

By simplifying the fraction we get:

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This is the simplified expression.

i hope it will help you!

4 0
2 years ago
What’s the area of the figure below?
TEA [102]

Answer:

482.5 ft^2.

Step-by-step explanation:

The figure can be regarded as a rectangle with a triangle cut out of one corner.

The base of the triangle = 25-20 = 5 in and the height = 20 - 13 = 7 in.

The required area = area of rectangle - area of triangle

= 25*20 - 1/2 * 5 * 7

= 500 - 17.5

= 482.5 ft^2

5 0
3 years ago
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