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2 years ago

-4y+4=10x standard form ik its 10x+4y=4 but you need to simplify more please help i need to get it in an hr

Mathematics

1 answer:

LuckyWell [14K]2 years ago

7 0

Answer:

x= (2/5)-(2y/5) (Im not sure if this is what you need but this is the answer simplified)

Step-by-step explanation:

Move all terms that don't contain x to the right side and solve.

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Find the slope of the line that goes through the given points (8,5) (6,7)

vazorg [7]

To find the slope of a line that goes through given points with known coordinates, you divide the subtraction of the y of the second point minus the y of the first point, by the subtraction of the x of the second point minus the x of the first point:
m = (yB-yA) / (xB-xA)

Let A(8,5) and B(6,7).
With yB = 7; yA = 5; xB = 6; xA = 8
m = (7-5) / (6-8)
m = 2/-2
m = -1

So the slope of the line that goes through the given points (8,5) and (6,7) is m = -1.

I've added a pic of the line with both points under the answer.

Hope this Helps! :)

4 0

3 years ago

Rewrite sin^4xtan^2x in terms of the first power of the cosine

Vikki [24]

$\sin ^4 x \tan ^2 x = \dfrac{\sin ^4 x\ \sin ^2 x }{\cos ^2 x} = \dfrac{\sin^6 x}{\cos ^2 x} = \dfrac{(\sin ^2 x)^3}{\cos^2 x}=\dfrac{(1-\cos^2 x)^3}{\cos^2 x}$

7 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

14 = 2y/5 what is the solution?

Gnom [1K]

Answer:

y=35

Step-by-step explanation:

y in (-oo:+oo)

14 = (2*y)/5 // - (2*y)/5

14-((2*y)/5) = 0

(-2/5)*y+14 = 0

14-2/5*y = 0 // - 14

-2/5*y = -14 // : -2/5

y = -14/(-2/5)

y = 35

5 0

3 years ago

D'Angelo bought two shirts at a special sale. He paid the full price of $19.99 for one shirt and got a second shirt of the same

Alex

Answer: $2.55

Step-by-step explanation:

He bought the second shirt for half off the first shirt's price. He therefore bought it at:

= 19.99/2

= $10.00

Total paid for both shirts is therefore:

= 19.99 + 10

= $29.99

Sales tax is 8.5% so he paid:

= 29.99 * 8.5%

= ‭2.54915‬

= $2.55

7 0

3 years ago