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4 years ago

ASAP On the coordinate plane below, quadrilaterals TRAP and HELP are similar to each other.

Mathematics

2 answers:

monitta4 years ago

7 0

Answer:

1:2.

Step-by-step explanation:

HP and TP are corresponding sides and their lengths are 2 and 4.

Thats a ratio of 1 :2.

The perimeter is also one dimensional so the ratio of the perimeters is also 1:2.

ipn [44]4 years ago

5 0

Answer:

1 : 2

Step-by-step explanation:

Since the quadrilaterals are similar then the ratios of corresponding sides are equal, that is

$\frac{HP}{TP}$ = $\frac{2}{4}$ = $\frac{1}{2}$

Thus the ratio of perimeters is also 1 : 2

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A health research institute collects information from 20 individuals on the number of years they have spent smoking cigarettes,

olga55 [171]

The correlation coefficient of the health research institute data measures the relationship between the age and the years of the smokers

The correlation coefficient is 0.53

<h3>How to calculate the correlation coefficient</h3>

The correlation coefficient (r) is calculated as:

$r = \frac{n(\sum xy) - \sum x \sum y}{\sqrt{[n \sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2}}$

Using the given parameters, we have:

$r = \frac{20 *8249 - 1257* 116}{\sqrt{[20 * 98823 - 1257^2][20 * 836 - 116^2}}$

Evaluate the exponents

$r = \frac{20 *8249 - 1257* 116}{\sqrt{[20 * 98823 - 1580049][20 * 836 - 13456}}$

Evaluate the products

$r = \frac{164980 - 145812}{\sqrt{[1976460 - 1580049][16720 - 13456}}$

Evaluate the differences

$r = \frac{19168}{\sqrt{[396411*3264}}$

Evaluate the product

$r = \frac{19168}{\sqrt{1293885504}}$

Evaluate the root

$r = \frac{19168}{35970.6}$

Evaluate the quotient

$r = 0.53$

Hence, the correlation coefficient is 0.53

Read more about correlation coefficient at:

brainly.com/question/1564293

8 0

3 years ago

Find the average value of the function f(t)=cos13(5t)sin(5t) f(t)=cos13⁡(5t)sin⁡(5t) on the interval [4,10].

Vilka [71]

The average value is given by

$\displaystyle\frac1{10-4}\int_4^{10}\cos^{13}5t\sin5t\,\mathrm dt$

Setting $y=\cos5t$ , you get $\mathrm dy=-5\sin5t\,\mathrm dt$ , and the integral becomes

$\displaystyle-\frac1{30}\int_{\cos20}^{\cos50}y^{13}\,\mathrm dy$
$=-\dfrac1{30}\dfrac{y^{14}}{14}\bigg|_{y=\cos20}^{y=\cos50}$
$=-\dfrac{\cos^{14}50-\cos^{14}20}{420}\approx-0.00145$