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4 years ago

Construct ∆ABC such that AB = 2.5cm, BC = 6cm and AC = 6.5cm. Measure angle B.

Mathematics

1 answer:

Goshia [24]4 years ago

3 0

Answer:

this is the best i could do

steps of construction :

draw a line segment with BC =6cm

with B as centre and 2.5 cm as radius draw an arc with compass

with c as centre and 6.5cm as radius draw another arc which cut the other arc .mark it as A.join A,B,C and triangle is formed. with help of protractor measure angle B

Step-by-step explanation:

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15 points!!! kinda rusty with triangles. scratch that. really rusty.

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Consider the attached figure. The height CD cuts the triangle exactly in half. This means that

$\overline{AD}=\overline{BD}=\dfrac{1}{2}\overline{AB}=10$

Moreover, since CD is the height of the triangle, we know that ACD is a right triangle. We know the hypothenuse AC to be 20 feet because it is a side of the triangle, and we just found out that AD is 10. We can use the pythagorean theorem to deduce

$\overline{CD}=\sqrt{\overline{AC}^2-\overline{AD}^2}=\sqrt{400-100}=\sqrt{300}$

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3 years ago

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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 w

irina1246 [14]

Answer:

A 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students who eat cauliflower

n = sample of students

p = population proportion of students who eat cauliflower

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

Now, in Agresti and Coull's method; the sample size and the sample proportion is calculated as;

$n = n + Z^{2}__(\frac{_\alpha}{2})$

n = $24 + 1.96^{2}$ = 27.842

$\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_) }{2} }{n}$ = $\hat p = \frac{2+\frac{1.96^{2} }{2} }{27.842}$ = 0.141

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ , $0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } }$ ]

= [0.012, 0.270]

Therefore, a 95% confidence interval for the proportion of students who eat cauliflower on Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95% confident that the proportion of students who eat cauliflower on Jane's campus is between 0.012 and 0.270.

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