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Sergeu [11.5K]
3 years ago
5

In a recent poll of 1100 randomly selected home delivery truck drivers, 26% said they had encountered an aggressive dog on the j

ob at least once. What is the standard error for the estimate of the proportion of all home delivery truck drivers who have encountered an aggressive dog on the job at least once? Round to the nearest ten-thousandth a) 0.0141 b) 0.0002 c) 0.1322 d) 0.0132
Mathematics
1 answer:
tankabanditka [31]3 years ago
5 0

Answer:

d) 0.0132

Step-by-step explanation:

In a sample of size n, with proportion p, the standard error of the proportion is:

SE_{p} = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.26, n = 1100

So

SE_{p} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.26*0.74}{1100}} = 0.0132

So the correct answer is:

d) 0.0132

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A typical construction worker makes $120 in eight hour work day in Arizona. How much do they make per hour?
son4ous [18]

Answer:

$15

Step-by-step explanation:

Simply divide 120 by 8 to get your answer.

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A) the greatest common factor is how to factors something

for this case x is the gcf

x(x^2 - 4) is the answer

B) Cannot be factored there is no common factor.

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Please someone help me to prove this..​
Pachacha [2.7K]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}

\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}

\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}

\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}

\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}

\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10

LHS = RHS:  cot 10 = cot 10   \checkmark

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Rewrite as: =(y^2-a)(1+x)-(y^2-a)b

Factor out the common term (y^2 - a): =(y^2-a)(1+x-b)

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