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adelina 88 [10]
3 years ago
8

Which of the following matches the graph? Y=3/4x Y=1/4x Y=1/3x Y=4x Y=4/3

Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0
The answer is y = 1/4x as the slope of the line is 1/4, found using the formula (y2 -y1) / (x2 - x1)
nika2105 [10]3 years ago
4 0

Answer:

B) y=1/4x

Step-by-step explanation:

Because the slope of this graph is 1/4. You can find this because slope=rise/run. On the graph, it's always 1 upward and 4 rightwards, therefore, the slope is 1/4. Also, the y-intercept is 0 in this case.

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Please help ASAP this is due tomorrow
scoray [572]

Answer:

c= hy−w /6y

Step-by-step explanation:This is what i got.

4 0
3 years ago
If two lines never intersect when graphed. Which of the following best describe the number of solution to this system of linear
serg [7]

Think of when If you cross two lines you get a point where they meet

A point is a solution.

So if the lines never cross, then they would have no point, or A, No Solution

3 0
3 years ago
Can you show me how you worked this: <br><br> Evaluate the expression.<br> 9! − 7!
Aloiza [94]
I hope this helps you



9.8.7!-7!


7! (72-1)

7!.71
3 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: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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

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

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

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}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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}

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

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}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

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}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
Write and equati9n of a l8ne that passes through (1,-2) and is perpendicular to -4x+7y=21
yarga [219]

Given:

A line passes through (1,-2) and is perpendicular to -4x+7y=21.

To find:

The equation of that line.

Solution:

We have, equation of perpendicular line.

-4x+7y=21

Slope of this line is

m_1=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}

m_1=-\dfrac{-4}{7}

m_1=\dfrac{4}{7}

Product of slope of two perpendicular lines is -1.

m_1\times m_2=-1

\dfrac{4}{7}\times m_2=-1

m_2=-\dfrac{7}{4}

Now, slope of required line is -\dfrac{7}{4} and it passes through (1,-2). So, the equation of line is

y-y_1=m(x-x_1)

where, m is slope.

y-(-2)=-\dfrac{7}{4}(x-1)

4(y+2)=-7(x-1)

4y+8=-7x+7

4y+7x=7-8

7x+4y=-1

Therefore, the equation of required line is 7x+4y=-1.

7 0
3 years ago
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