All categories

3 years ago

One polygon has a side of length 3 feet. A similar polygon has a corresponding side of length 9 feet. The ratio of the perimeter

of the smaller polygon to the larger is 3:1 1:6 1:3

Mathematics

1 answer:

shusha [124]3 years ago

6 0

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill \\\\ \cfrac{\textit{small polygon}}{\textit{large polygon}}\qquad \qquad \cfrac{3}{9}\implies \cfrac{1}{3}\implies \stackrel{ratio}{1:3}$

You might be interested in

The results of Accounting Principals’ latest Workonomix survey indicate the average American worker spends $1092 on coffee annua

Nataly_w [17]

Answer:

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

According to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average, 47.74% probability that a person between 35 and 44 consume more than the average and 50% probability that a person older than 45 consume more than the average.

Step-by-step explanation:

The mean for each sample is

$\bar x=\frac{\sum_{k=1}^{10}x_k}{10}$

where the $x_k$ are the data corresponding to each group

The variance is

$s^2=\frac{\sum_{k=1}^{10}(\bar x-x_k)^2}{9}$

and the standard deviation is s, the square root of the variance.

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

Let's compare these averages against the general media established of $1,092 by using the corresponding z-scores

$z=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

$\bar x$ = mean of the sample

$\mu$ = established average

s = standard deviation of the sample

n = size of the sample

z-score of Group 18-34 years old

$z=\frac{1041.625-1092}{696.635/\sqrt{10}}=-0.2286$

The area under the normal curve N(0;1) between -0.2286 and 0 is 0.0904. So according to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average.

z-score of Group 35-44 years old

$z=\frac{1359-1092}{422.5491/\sqrt{10}}=2.0019$

The area under the normal curve N(0;1) between 0 and 2.0019 is 0.4774. So according to the sample there is 47.74% probability that a person between 35 and 44 consume more than the average.

z-score of Group 45 and older

$z=\frac{1414.375-1092}{135.2489/\sqrt{10}}=7.5375$

The area under the normal curve N(0;1) between 0 and 7.5375 is 0.5. So according to the sample there is 50% probability that a person older than 45 consume more than the average.

3 0

4 years ago

X-c= y-rd, for x . . . . . . . . . . . . . .

ANTONII [103]

Answer:

x = y - rd + c

Step-by-step explanation:

x - c = y -rd

x = y - rd + c

3 0

3 years ago

Find the quotient 1/5 divided by 2