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Butoxors [25]
3 years ago
13

A coordinate grid is mapped onto a video game screen, with the origin at the lower left corner. The game designer programs a tur

tle to move along a linear path that passes through the points (0, 0) and (10, 8).
The designer also programs a bird with a path that can be modeled by a quadratic function. The bird starts at the vertex of the path at (0, 20) and passes through the point (10, 8).

What is the slope of the line that represents the turtle's path?
Mathematics
2 answers:
Tanya [424]3 years ago
4 0
Is the second paragraph meant to throw you off??? Because the answer to the question does not rely on the information based from the second paragraph...

Anyway, we know that the turtle started from (0, 0) and moved to (10, 8). Let's assume that (0, 0) is (x1, y1), and that (10, 8) is (x2, y2) for the sake of calculations.

To find how far the turtle moved to the right (along the x-axis) we take x1 from x2:

x2 - x1 = 10 - 0
= 10 = x2
***this is the "run"

To find how far the turtle moved upwards (along the y-axis) we take y1 from y2:

y2 - y1 = 8 - 0
= 8 = y2
***this is the "rise"

Notice that when something moves away from the origin the distance it traveled along the x-axis and y-axis is the same as its final position (or coordinates to where it traveled to).

Therefore, the,

slope = rise / run
= x2 / y2
= 10 / 8
= 5 / 4 (preferable for use)
= 1.25
faltersainse [42]3 years ago
4 0

Answer:

The answer is 0.8

Step-by-step explanation:

I'm doing the assignment now.

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Marla swims twice a week. her equipment cost her $32.85 and she has a membership to the pool for $35 plus #3.50 per visit. she c
evablogger [386]
Around $634.85 per year

Because:

$32.85+$35(12 months)+$3.5(52 weeks in a year)= Total cost per year
                   $420                       $182

        $32.85 + $420 + $182 = $634.85 per year
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
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Ratling [72]
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8 0
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What is the gcf of 56 and 14
IgorC [24]

Answer:

14

Step-by-step explanation:

14/14=1

56/14=4

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Answer:

35-22=x

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Step-by-step explanation:

Include units if necessary:

$35-$22=$13

8 0
3 years ago
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