<img src="https://tex.z-dn.net/?f=%20%20%5Cfrac%7B%20%5Csqrt%7B%204%7D%20%7D%7B3%20%5Csqrt%7B25%7D%20%7D%20" id="TexFormula1" ti

All categories

2 years ago

tle=" \frac{ \sqrt{ 4} }{3 \sqrt{25} } " alt=" \frac{ \sqrt{ 4} }{3 \sqrt{25} } " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

kirza4 [7]2 years ago

5 0

Answer:

7.5........................

You might be interested in

1. Find 1/8 x 2/3 *

mash [69]

1. is 1/12. 2. is 1 3/7

8 0

3 years ago

Probability allows one to measure effectively the risks in selecting one alternative over the others.

kotegsom [21]

The answer to your question is “false”

6 0

3 years ago

HEEELLLPPP!!!!!!!!! PLEEAASEEE!!!!!!!!

Ksivusya [100]

Answer:

Step-by-step explanation:

$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$

4 0

2 years ago

A) 20 milligrams of the drug lincomycin is to be given for each kilogram of a person's weight the drug is be mixed with 250 cc o

krok68 [10]

Answer:

you need to convert pounds to kilograms. 196 pounds is 88.904 kilograms. The directions tell you that 20 milligrams of the drug is to be administered for each kilogram of a person's weight. Now you need to multiply 88.904 by 20. That should give you 1778.08 milligrams of lincomycin. You would then mix 250 cc's of saline solution to the 1778.08 milligrams of lincomycin; but I think its just asking you how much lincomycin to administer.

at what rate per minute should the 250 cc solution be adminstered

4 0

3 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use of one particular cust

Fiesta28 [93]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

$321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320$

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}$

STEP 3: Calculate the mean

$\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}$

STEP 4: Calculate the Standard deviation

$\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}$

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

$\begin{gathered} \text{IQR}=Q_3-Q_1 \\ \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\ \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}$

STEP 6: Find the Interquartile Range

$\begin{gathered} IQR=Q_3-Q_1 \\ \text{IQR}=479.5-363.5 \\ \text{IQR}=116 \end{gathered}$

Hence, the interquartile range of the data is 116

3 0

1 year ago