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4 years ago

Consider triangle ABC and triangle ACD in the diagram below. Which of the following complete the proof above?

Mathematics

1 answer:

Nata [24]4 years ago

7 0

Given:

AC is perpendicular bisector of BD

AB ≅ AD

To prove:

ΔABC ≅ ΔACD

Solution:

In triangle ABC and triangle ACD,

Step 1: Given

AC is perpendicular bisector of BD and

AB ≅ AD (Side)

Step 2: By perpendicular bisector definition,

BC ≅ CD (Side)

Step 3: By reflexive property,

AC ≅ AC (Side)

Step 4: By SSS congruence criterion,

ΔABC ≅ ΔACD

Hence proved.

You might be interested in

A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat

Marizza181 [45]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

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

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 $\alpha=1-0.90=0.1$ 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 $ME =\pm 0.04$ 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 assume that a prior estimation for p would be $\hat p =0.5$ since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

5 0

3 years ago

How many books must be chosen from among?

Strike441 [17]

You have to give more explanation

5 0

3 years ago

Who is farthest from the school

postnew [5]

Using the law of cosines:

The grocery store to the school:

Distance = √(7^2 + 10^2 - 2*7*10*cos(100)

Distance = 13.16 miles

The movie store to the school:

Distance = √(7^2 + 10^2 - 2*7*10*cos(120)

Distance = 14.80 miles

The friend is further away.

7 0

4 years ago

What is the area of a oblong with a base of 7cm and a height of 4 cm

mariarad [96]

Answer: 7*4 is 28 square centimeters...

7 0

2 years ago

1).Calculate T, when sample mean is 120, population mean is 100, standard deviation is 20 and smaple size is 10 using levels of

Sonbull [250]

Answer:

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$

$p_v =2*P(t_{9}>3.16)=0.012$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=120$ represent the sample mean

$s=20$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=10-1=9$

Conclusion

Since is a tao tailed test the p value would be:

$p_v =2*P(t_{9}>3.16)=0.012$

8 0

3 years ago

Read 2 more answers