between what two numbers would 9 be located on a number line? a. 10 and 11 b. 7 and 8 c. 8 and 10 d. 9 and 10
2 answers:
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1)
∠BAC = ∠NAC - ∠NAB = 144 - 68 = 76⁰
AB = 370 m
AC = 510 m
To find BC we can use cosine law.
a² = b² + c² -2bc*cos A
|BC|² = |AC|²+|AB|² - 2|AC|*|AB|*cos(∠BAC)
|BC|² = 510²+370² - 2*510*370*cos(∠76⁰) =
|BC| ≈ 553 m
2)
To find ∠ACB, we are going to use law of sine.
sin(∠BAC)/|BC| = sin(∠ACB)/|AB|
sin(76⁰)/553 m = sin(∠ACB)/370 m
sin(∠ACB)=(370*sin(76⁰))/553 =0.6492
∠ACB = 40.48⁰≈ 40⁰
3)
∠BAC = 76⁰
∠ACB = 40⁰
∠CBA = 180-(76+40) = 64⁰
Bearing C from B =360⁰- 64⁰-(180-68) = 184⁰
4)
Shortest distance from A to BC is height (h) from A to BC.
We know that area of the triangle
A= (1/2)|AB|*|AC|* sin(∠BAC) =(1/2)*370*510*sin(76⁰).
Also, area the same triangle
A= (1/2)|BC|*h = (1/2)*553*h.
So, we can write
(1/2)*370*510*sin(76⁰) =(1/2)*553*h
370*510*sin(76⁰) =553*h
h= 370*510*sin(76⁰) / 553= 331 m
h=331 m
Answer:
The sample 2 has a lowest value of SE corresponding to the least sample variability.
Step-by-step explanation:
As the value of the sample means and standard deviations are not given, as similar question is found online from which the values of data is follows
The data is as attached with the solution. From this data
Sample 1 has a mean of 34 and a SE of 5
Sample 2 has a mean of 30 and a SE of 2
Sample 3 has a mean of 26 and a SE of 3
Sample 4 has a mean of 38 and a SE of 5
As per the measure of the sample variability is linked with the value of SE or standard error. Which is lowest in the case of sample 2 .
So the sample 2 has a lowest value of SE corresponding to the least sample variability.
