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Dmitriy789 [7]
4 years ago
14

lesson 6 unit 7 triangles sample work for geometry (I just need answers to the highlighted ones) 15pts

Mathematics
1 answer:
RSB [31]4 years ago
7 0

Answer:

2.  Circumcenter is (-2, 1)

4.  Orthocenter is (5, 0)

6. II ∠DOS and ∠CAT are vertical and III ∠DOS and ∠CAT are adjacent

8.  m∠PBC = 30°

10.  KV, VM, KM

12.  DC is greater

14.  x = 12

Step-by-step explanation:

2.  Equation of the perpendicular bisector of AB is x = -2 and the equation of the perpendicular bisector of BC is y = 1.

Circumcenter is the point of intersection of the perpendicular bisectors of the sides of the triangle.

Hence, circumcentre is (-2, 1).

4.  Orthocenter is the point of intersection of the altitudes. Here, AB and BC are two altitudes and B is their point of intersection.

Hence, the orthocenter is (5, 0).

6.  Adjacent angles are angles separated by a line and vertical angles are angles vertically opposite to each other.

Hence, statements II and III contradict to each other.

8.  If P is equidistant from all the three sides, then P is the center of the in-circle and hence it is the incenter of the triangle.

If P is the incentre, then PB is the internal bisector.

Therefore, ∠PBC = \frac{1}{2} ∠ABC = 30°

10.  From the triangle sum property, ∠ M = 180 - (110 + 40) = 30.

Since, the side opposite to the smallest is the shortest, the correct order is:

KV, VM, KM

12.  Clearly, ∠ ABD is acute. Therefore, ∠ DBC > ∠ ABD.

Since, side opposite to larger angle is greater, DC is greater.

14.  Let the sides of the top triangle be p and q. So, two sides of whole triangle are 2p and 2q.

Now, if we compare the top small triangle and the whole triangle, they are similar by SAS similarity since,

the top angle is common and \frac{p}{2p}=\frac{q}{2q}=\frac{1}{2}.

Since, corresponding parts of similar triangles are proportionate,

\frac{3x}{5x+12} =\frac{1}{2}

6x = 5x + 12

6x - 5x = 12

x = 12

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a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation  

n=2700 represent the random sample taken    

X represent the people in England who are deficient in Vitamin D

\hat p=0.2 estimated proportion of England people who are deficient in Vitamin D

\alpha=0.02 represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)    

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

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

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182

0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

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