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ipn [44]
2 years ago
13

F(x)=3x-7 input is 3

Mathematics
1 answer:
Triss [41]2 years ago
5 0

Answer:

The answer is f(x)=2

Step-by-step explanation:

When the input is given it means that the x is substituted by that number.

so 3x3-7.

3x3-7

9-7=2.

The f of x is equivalent to 2.

I hope this helps :D

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Answer:

It was ordinary

Step-by-step explanation:

7 0
3 years ago
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6. Complete the two-column proof.
olya-2409 [2.1K]

Answer:

Given: \frac{x}{6} + 2 = 15                             ......[1]

To prove : x =78

Subtraction property states that you subtract the same number to both sides of an equation.

Subtract 2 from both sides of an equation [1];

\frac{x}{6} + 2 - 2= 15 -2

Simplify:

\frac{x}{6} = 13                                             ......[2]

Multiplication property states that you multiply the same number to both sides of an equation.

Multiply 6 to both sides of an equation [2];

\frac{x}{6} \times 6 = 13 \times 6

Simplify:

x = 78                      proved!

Statement                                           Reason

1.  \frac{x}{6} + 2 = 15                 Given

2.  \frac{x}{6} = 13                 Subtraction property of equality

3. x = 78                              Multiplication property of equality

8 0
3 years ago
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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
Please help on number 15
borishaifa [10]
....................................................................
6 0
3 years ago
0.4= Blankx10 please answer quick
Marta_Voda [28]

Answer:

1, 0.4

2, 0.004

3, 0.04

Step-by-step explanation:

hope this helps.      :)

6 0
3 years ago
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