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leva [86]
3 years ago
10

Using the right triangle below find the cotangent of angle A

Mathematics
1 answer:
pogonyaev3 years ago
7 0

Answer:

option d is correct

Step-by-step explanation:

cot=b/p

p=2

b=2\sqrt{3}

so b/p=2\sqrt{3}/2=\sqrt{3}

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HK bisects GHJ. Find GHK and KHJ. Angle JHG measures 161
lapo4ka [179]

Answer: m∠GHK=80.5,

m∠KHJ=80.5

Step-by-step explanation:

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3 years ago
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What type of line is perpendicular to a linear equation that is not a function
enyata [817]

Answer:

Determine whether lines are parallel or perpendicular given their equations; Find equations of . We can begin by using point-slope form of an equation for a line. No. For two perpendicular linear functions, the product of their slopes is –1.

Step-by-step explanation:

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3 years ago
One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe
sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% 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 people who disagree with the statement = \frac{1812}{2625} = 0.69

           n = sample of respondents = 2625

           p = population proportion of people who disagree with statement

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                   of significance are -2.58 & 2.58}  

P(-2.58 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.58) = 0.99

P( -2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [ \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

= [ 0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } , 0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } ]

 = [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

5 0
3 years ago
Plz help it’s due tonight!!!
Nutka1998 [239]
D. Is the ans
Please mark me as Brainliest
8 0
2 years ago
CONSTRUCTION You are asked to copy a segment CD, construct a segment bisector by paper folding, and label the
lilavasa [31]

The steps on the construction of a segment bisector by paper folding, and label the midpoint M is given below.

<h3>What are the steps of this construction?</h3>

1. First, one need to open a Compass so that it is said to be more than half the length of the said segment.

2. Without altering it, with the aid of the compass, do  draw an art above and also below the said line segment from one of the segment endpoints.

3. Also without altering it and with use the compass, do draw another pair of arts from the other and points. One arc will be seen above the segment and the other or the second arc will be seen below.

4. Then do draw the point of intersection that is said to exist between the pair of arts below the line segment and also in-between the pair of arts as seen  below the line segment

5. Lastly, do make use of a straight edge to link the intersection points between the both pair of arts.

Learn more about segment bisector from

brainly.com/question/24736149

#SPJ1

5 0
2 years ago
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