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3 years ago

A scale distance of 3 centimeters on a map represents an actual distance of 51 feet.

2 answers:

8 0

In this case, the best way to work this out is to find what 1cm is equal to. In this case, you have to divide 51/3, giving you 17. You know that 17 feet. You can now divide 187/17, giving you 11. Therefore, 11cm is equal to 187 feet. Hope this helps

Vanyuwa [196]3 years ago

7 0

If the distance is 51 feet in 3 cm
if the destance is 187 feet ..?cm
51 ....3
187...? =3 * 187/51
=3*3.58
=10.74 cm (hope it help)

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What type of transformation takes the graph of

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2 years ago

A random sample of 10 employees of a company was selected to estimate the mean one-way commute time for all employees at the com

liberstina [14]

Answer:

The margin of error is of $2.262\frac{6}{\sqrt{10}} = 4.29$

Step-by-step explanation:

We have the standard deviation for the sample. So the T-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.262

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.262\frac{6}{\sqrt{10}} = 4.29$

In which s is the standard deviation of the sample and n is the size of the sample.

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2 years ago

The sum of √12 and √3 is

forsale [732]

Step-by-step explanation:

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2 years ago

Simplify sin Θ/√1-sin^2 Θ

dedylja [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2862279

_______________

Simplify the trigonometric expression.

Call the expression E:

$\mathsf{E=\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}\qquad\qquad(but~~1-sin^2\,\theta=cos^2\,\theta)}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\sqrt{cos^2\,\theta}}}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\left|cos\,\theta\right|}}}$

$\mathsf{E}=\left\{\!\begin{array}{rl}\mathsf{\dfrac{sin\,\theta}{cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta>0}\\\\\mathsf{\dfrac{sin\,\theta}{-\,cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta$

So basically E is equivalent to $\pm\,\mathsf{tan\,\theta},$ where the sign depends on which quadrant $\theta$ lies. Shortly,

$\mathsf{\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}}=\left\{\!
\begin{array}{rl}
\mathsf{tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~1^{st}~or~the~4^{th}~quadrant}\\\\
\mathsf{-\,tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~2^{nd}~or~the~3^{rd}~quadrant}
\end{array}$

I hope this helps. =)

Tags: <em>simplify trigonometric trig expression sine cosine tangent sin cos tan absolute value modulus trigonometry</em>

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