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3 years ago

Convert y=-1.4x+9.6 to standard form

Mathematics

1 answer:

pychu [463]3 years ago

6 0

Answer:

14x-10y=-96

Step-by-step explanation:

we know that

The equation of the line in standard form is equal to

Ax+By=C

where

A is a positive integer

B and C are integers

we have

y=1.4x+9.6

Multiply by 10 both sides

10y=14x+96

isolate the variable x and y (Remember that A must be positive)

14x-10y=-96 -----> standard form

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Write an inequality to represent the situation. <br> A cheese pizza costs at least $5.

Nady [450]

<
- 5 dollars, the V thing sits on top of the line and 5 comes after signifying that it could be 5 dollars or more

8 0

3 years ago

Read 2 more answers

Two trains, Train A and Train B weigh a total of 349 tons. Train A is heavier than Train B. The difference in their weights is 2

Ainat [17]

Answer:

Train A = 299 tons Train B = 50 tons

Step-by-step explanation:
Train A = x

Train B = y

x+y = 349

x= y +249

Plug in one equation into another

y+249+y= 349

Simplify

2y+249=349

Subtract 249 from both sides

2y=100

Divide both sides by 2

y=50

Plug it into the x equation to solve for x

x+50=349

-50 from both sides

x=299

3 0

2 years ago

Evaluate the function f(x) at the given numbers (correct to six decimal places).

statuscvo [17]

Answer:

The values of given function are shown in the below table.

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2-5x}{x^2-x-20}$

Simplify the given function.

$f(x)=\frac{x(x-5)}{x^2-5x+4x-20}$

$f(x)=\frac{x(x-5)}{x(x-5)+4(x-5)}$

$f(x)=\frac{x(x-5)}{(x+4)(x-5)}$

Cancel out the common factor.

$f(x)=\frac{x}{x+4}$

Substitute x=5.5 in the above equation.

$f(5.5)=\frac{5.5}{5.5+4}$

$f(5.5)=\frac{5.5}{9.5}$

$f(5.5)=0.57894736842$

$f(5.5)\approx 0.578947$

Similarly find the value for all values of x.

The values of given function are shown in the below table.

8 0

3 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago

Y=1/2 linear or non linear

vladimir1956 [14]

Answer:

linear

Step-by-step explanation:

this is the graph of y = 0x + 0.5 simplified to y = 0.5
you can type the equation of a graph in desmos and see what it looks like

8 0

3 years ago