This is what I got for it if it helps
Two people are standing on opposite sides of a hill. Person A makes an angle of elevation of 65° with the top of the hill and pe
rson B makes an angle of elevation of 80° with the top of the hill. The two people are standing 45 feet from each other. What is the distance from person B to the top of the hill?
2 answers:
Answer:
71.10 feet
Step-by-step explanation:
To solve this you have to remember the laws of the triangles and the law of sines, this states that every triangle´s sum of the inner angles are 180º
This means that if angle A is 65º and angle B is 80º angle C which would be the angle created on the top oof the mountain byt the lines that conect the top of the mountain with person A and B would be 35º.
Now the law of sines goes like this:
If you put the values of the different angles you´d get:
When you clear the equation it will end up like this:
x=
x= 71.1
See the picture to better understand the problem
we know that
in the triangle ABC
∠A+∠B+∠C=180°
find ∠C
∠C=180-[80+65]------> ∠C=35°
Applying the law of sines
AB/sin C=CB/sin A
solve for CB
CB=AB*sin A/sin C-----> CB=45*sin 65/sin 35-----> CB=71.10 ft
the answer is
<span>the distance from person B to the top of the hill is 71.10 feet</span>
we know that
in the triangle ABC
∠A+∠B+∠C=180°
find ∠C
∠C=180-[80+65]------> ∠C=35°
Applying the law of sines
AB/sin C=CB/sin A
solve for CB
CB=AB*sin A/sin C-----> CB=45*sin 65/sin 35-----> CB=71.10 ft
the answer is
<span>the distance from person B to the top of the hill is 71.10 feet</span>
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Answer:
Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
NyQuil Robitussin Triaminic Total
No relief 11 13 9 33
Some relief 32 28 27 87
Total relief 7 9 14 30
Total 50 50 50 150
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the three remedies
H1: There is a difference in the three remedies
The level os significance assumed for this case is
The statistic to check the hypothesis is given by:
The table given represent the observed values, we just need to calculate the expected values with the following formula
And the calculations are given by:
And the expected values are given by:
NyQuil Robitussin Triaminic Total
No relief 11 11 11 33
Some relief 29 29 29 87
Total relief 10 10 10 30
Total 50 50 50 150
And now we can calculate the statistic:
Now we can calculate the degrees of freedom for the statistic given by:
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.81,4,TRUE)"
Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.