9. Karl is putting a frame around a rectangular photograph. The photograph is

EAB and DCB are two right triangles. The figure has BED≅ BDE. Point B is the midpoint of segment AC. Prove: EAB≅ DCB

aleksklad [387]

See the attached picture.

you are given that ABCE is an isosceles trapezoid. 

you are given that AB is parallel to EC. 

this means that AE is congruent to BC. 

you are given that AE and AD are congruent. 

triangle EAD is an isosceles triangle because AE and AD are congruent. 

this means that angle 1 is equal to angle 3. 

since angle 1 is equal to angle 2 and angle 3 is equal to angle 1, then angle 3 is also equal to angle 2. 

this means that AD and BC are parellel because their corresponding angles (angles 3 and 2) are equal. 

since AB is parallel to EC and DC is part of the same line, than AB is parallel to DC. 

you have AB parallel to DC and AD parallel to BC. 

if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. 

that might be able to do it,depending on whether all these statements are acceptable without proof. 

they are either postulates or theorems that have been previously proven. 

if not, then you need to go a little deeper and prove some of the statements that you used.. 

here's my diagram.

this is not a formal proof, but should give you some ideas about how to proceed. 

you can also prove that angle 4 is equal to angle 2 because they are alternate interior angles of parallel lines. 

you can also prove that angle 6 is equal to angle 5 because they are alternate interior angles of parallel lines.

5 0

1 year ago

The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of

grandymaker [24]

Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

Let $\mu$ = mean gross annual incomes of certified welders

So, Null Hypothesis, $H_0$ : $\mu$ = $20,000

Alternate hypothesis, $H_A$ : $\mu \neq$ $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean income

$\sigma$ = population standard deviation = $2,000

n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

3 0

2 years ago

Yoko evaluates 7 divide 1/6 by using a related multiplication expression. Which multiplication expression should she use?

Olin [163]

$\bf 7\div \cfrac{1}{6}\implies 7\cdot \cfrac{6}{1}$

4 0

1 year ago

Read 2 more answers

A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa

sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) $p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference $d_i=A_i-B_i$ and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_A- \mu_B = 0$

Alternative hypothesis: $\mu_A -\mu_B \neq 0$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0

2 years ago

at the community savings bank it takes a computer 30 minutes to process and print payroll checks when the second computer is use

s2008m [1.1K]

Well im not to sure about yours but mine say the answer is A

6 0

2 years ago