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3 years ago

3x+2y=3y-2 line 1 x+y=10 line 2 solve for x/y

Mathematics

1 answer:

GrogVix [38]3 years ago

3 0

Answer:

x/y=1/4

Step-by-step explanation:

3x+2y=3y-2

3x=3y-2y-2

3x=y-2

y=3x+2

x+y=10

x+3x+2=10

4x+2=10

4x=10-2

4x=8

x=8/4

x=2

2+y=10

y=10-2

y=8

x/y=2/8=1/4

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Answer:

a>-2

Step-by-step explanation:

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What's 3.82e+7 in standard form

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If you simplify, it equals 10.82e.

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Read 2 more answers

When 200 gallons of oil were removed from a tank, the volume of oil left in the tank was 3/7 of the tank's capacity. what was th

ki77a [65]

Answer:

Step-by-step explanation:

"When 200 gallons of oil were removed from a tank" algebraically looks like this:

V - 200.

"...the volume of oil left in the tank was 3/7 of the tank's capacity" algebraically looks like this:

3/7(V)

Therefore, the equation is

V - 200 = 3/7(V)

Begin by multiplying both sides by 7:

7(V - 200) = 3V and

7V - 1400 = 3V so

-1400 = -4V so

V = 350 gallons

That's if the volume of the oil in the tank was 3/7 of the tank's capacity.

For the other part of the problem, we set up the equation almost the same, except the 3/7 is a 1/2:

V - 200 = 1/2(V)

Multiplying both sides by 2 gives you

2(V - 200) = V and

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3 years ago

Find the solutions of the quadratic equation -9c² +2 +3 = 0.

Tresset [83]

Given that,

An equation : -9c² +2c +3 = 0

To find,

Find the value of x.

Solution,

We have, -9c² +2c +3 = 0

We can solve it using the formula as follows :

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, a = -9, b = 2 and c =3

Put the values,

$c=\dfrac{-2\pm \sqrt{(2)^2-4(-9)(3)}}{2(-9)}\\\\c=\dfrac{-2+\sqrt{(2)^{2}-4(-9)(3)}}{2(-9)}, \dfrac{-2-\sqrt{(2)^{2}-4(-9)(3)}}{2(-9)}\\\\c=-0.47, 0.69$

So, the solution are -0.47 and 0.69.

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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