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3 years ago

Which equation can be used to solve for "X"?

Mathematics

1 answer:

ipn [44]3 years ago

5 0

Answer:

im going to tell the teacher for cheating

Step-by-step explanation:

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7 times as much as the sum of 1/3 and 4/5

Vlada [557]

$7( \frac{1}{3} + \frac{4}{5} )$

$=7( \frac{1(5)}{3(5)} + \frac{4(3)}{5(3)} )$

$=7( \frac{5}{15} + \frac{12}{15} )$

$=7( \frac{17}{15})$

$=\frac{7}{1} ( \frac{17}{15})$

$=\frac{119}{15}$

$= 7\frac{14}{15}$ ≈ 7.93

4 0

3 years ago

Read 2 more answers

What is the mean salary of a salesperson at the company

Whitepunk [10]

<span>a fixed regular payment, typically paid on a monthly or biweekly basis but often expressed as an annual sum, made by an employer to an employee, especially a professional or white-collar worker.</span>

7 0

3 years ago

WILL GIVE BRAINLIST SOON AS I CAN SOMEBODY PLEASE HELP ME

fgiga [73]

Answer:

A= 75.06 Sq Cm

Step-by-step explanation:

Please refer to the picture attached with this.

In ΔABD

AB = 8.5 , AD = 6 and say BD = x

Applying Pythagoras theorem in this triangle

Pythagoras theorem says

$a^2+b^2=c^{2}$

where c is the side opposite to the right angle, and b and a are the other two sides of the right angled triangle.

$x^2+6^2=8.5^{2}$

$x^2=72.25-36$

$x^2=36.25$

$x=6.02$

Hence side $a = x +19$

a = 6.02+19

a = 25.02

Now we apply the formula for area of a triangle which is given as

$A=\frac{1}{2}\times b \times h$

Where b is the base , here we have base as a = 25.02

h is the height , here h = 6

Putting these values in Formula we get

$A=\frac{1}{2}\times 25.02 \times 6$

$A= 25.02 \times 3$

$A=75.06$

Hence Area of the triangle is 75.06 sq cm

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3 years ago

At the newest animated movie, for every 999 children, there are 444 adults. There are a total of 393939 children and adults at t

bogdanovich [222]

27 children at the movie

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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