What is the interquartile rang of the following data set? 9,30,2,48,42,18,81,5,55,73,11

Mathematics

1 answer:

ycow [4]2 years ago

6 0

30 is the real answer

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A person places $290 in an investment account earning an annual rate of 2.2%, compounded continuously. Using the formula V = Pe^

Tomtit [17]

Answer:

Step-by-step explanation:

Given that the principal p= $290

rate r= 2.2% 2.2/100 =0.022

time t= 18years

by applying the expression

$V = Pe^{rt}$

We have

$V = 290e^{0.022*18}\\\\V=290e^{0.396}\\\\V=290*1.48586931755\\\\V=$430.90$

Hence after 18years the money in the account will be $430.90

8 0

3 years ago

39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t

Katyanochek1 [597]

Write tan in terms of sin and cos.

$\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}$

Recall that

$\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1$

Rewrite and expand the given limand as the product

$\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)$

Then using the known limit above, it follows that

$\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}$