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nevsk [136]
3 years ago
11

PLEASE HELP NOW 25pts!! O 60x10

Mathematics
1 answer:
alex41 [277]3 years ago
3 0

Answer:

B.

Step-by-step explanation:

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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
X-1/x-2+x+3/x-4=2/(x-2).(4-x)
Varvara68 [4.7K]

The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

According to the statement

we have given that the equation and we have to prove that the given answer is a correct answer for those equivalent equation.

So, The given expression are:

\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}

And we have to prove the answer.

So, For this

\frac{x-1}{x-2} +\frac{x+3}{x-4}

\frac{({x-1}) ({x-4}) +({x+3})({x-2})} {(x-2) (x-4)}

Then the equation become

\frac{x^{2} -4x -x +4 + x^{2} -2x + 3x -6 }{(x-2) (x-4)}

Now solve it then

2x^{2} - 4x -2 / (x-2) (x-4)

Now take 2 common from answer then equation become

\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}

Hence proved.

So, The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

Learn more about equations here

brainly.com/question/2972832

#SPJ1

3 0
1 year ago
Select all the statements that are true about the linear equation. y = 4x - 3
Andre45 [30]
The answer is B. Have a nice day
8 0
3 years ago
Read 2 more answers
Question 7: Which of the following statements is true?
yawa3891 [41]

Problem 7)

The answer is choice B. Only graph 2 contains an Euler circuit.

-----------------

To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.

With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.

================================================

Problem 8)

The answer is choice B) 5

-----------------

Work Shown:

abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10

101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10

101 base 2 = (1*4 + 0*2 + 1*1) base 10

101 base 2 = (4 + 0 + 1) base 10

101 base 2 = 5 base 10


4 0
2 years ago
NEED HELP ASAP!!!!!!
Vinvika [58]

Answer:

2.3 hours because 18/60=.3 and add the 2 hours

7 0
2 years ago
Read 2 more answers
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