2:4 14:20 proportion? show work thx

1 answer:

4 0

$2:4=\dfrac{2}{4};\ 14:20=\dfrac{14}{20}\\\\\dfrac{2}{4}=\dfrac{14}{20}\ \ \ \ |cross\ multiply\\\\2\cdot20=4\cdot14\\\\40=56\ \ \ |FALSE$

Answer: It's not proportion.

You might be interested in

(-2/3, sqrt5/3) is a point on a unit circle. Find the cosine, cosecant, and sine of the angle.

Nana76 [90]

Look at the picture.

$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{y}{r}}=\dfrac{r}{y}$

We have the right triangle x, y and r. From the Pythagorean theorem we have:

$r^2=x^2+y^2\to r=\sqrt{x^2+y^2}$

We have the point

$\left(-\dfrac{2}{3};\ \dfrac{\sqrt5}{3}\right)$

Substitute:

$r=\sqrt{\left(-\dfrac{2}{3}\right)^2+\left(\dfrac{\sqrt5}{3}\right)^2}\\\\r=\sqrt{\dfrac{4}{9}+\dfrac{5}{9}}\\\\r=\sqrt{\dfrac{9}{9}}\\\\r=1$

$\csc\theta=\dfrac{1}{\frac{\sqrt5}{3}}=\dfrac{3}{\sqrt5}=\dfrac{3\cdot\sqrt5}{\sqrt5\cdot\sqrt5}=\boxed{\dfrac{3\sqrt5}{5}}$

8 0

3 years ago

A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/

xz_007 [3.2K]

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$