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3 years ago

Graph the line represented by the equation.

Mathematics

1 answer:

harina [27]3 years ago

5 0

Answer:

Step-by-step explanation:

first you need to put the equation in y=mx+b form

y-3=(-1/2)(x+4)

y-3= (-1/2x)+4

y=-1/2x+1

then you just graph

the y intercept is 1 and the slope is -1/2 so it’s going to be a negative graph and it starts from 1 on the y axis

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Question 3 of 10 Evaluate: (4/5) 1/2 O A. 2/25 O B. 2/5 O C.4:5 O D.4/25

Natasha2012 [34]

Answer:

2/5

Step-by-step explanation:

5 0

2 years ago

Choose yes or no to tell whether the expression represent a 20% discount off the price of an item that originally cost d dollars

Vaselesa [24]

Answer: B. d - 0.2

Step-by-step explanation: 20%= 20/100= 0.2

Which means = d-0.2

Hope this helped.

8 0

3 years ago

Read 2 more answers

8n-13=13-8n can someone answer with steps, please?

NARA [144]

Answer:

1.625

Step-by-step explanation:

Add 13 to both sides

8n = 26 - 8n

Add 8n to both sides

16n = 26

Divide both sides by 16

Your answer is 1.625

Hope I helped :)

Please consider Brainliest for me :)

(unless the other person does a better job)

(Which they probably will) :)

4 0

2 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

Part a at what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a di

loris [4]

3.01 m/s
This is a simple projectile calculation. What we want is a vertical velocity such that the time the droplet spends going up and going back down to the surface exactly matches the time the droplet takes to travel horizontally 0.800 meters. The time the droplet spends in the air will is:
V*sqrt(3)/2 ; Vertical velocity.
(V*sqrt(3)/2)/9.8 ; Time until droplet reaches maximum height
(V*sqrt(3))/9.8 ; Double that time for droplet to fall back to the surface.

The droplet's horizontal velocity will be:
V/2.

So the total distance the droplet travels will be:
d = (V*sqrt(3))/9.8 * V/2
d = V^2*sqrt(3)/19.6

Let's substitute the desired distance and solve for V
d = V^2*sqrt(3)/19.6
0.8 = V^2*sqrt(3)/19.6
15.68 = V^2*sqrt(3)
15.68/sqrt(3) = V^2
15.68/1.732050808 = V^2
3.008795809 = V
So after rounding to 3 significant figures, the archerfish needs to spit the water at a velocity of 3.01 m/s
Let's verify that answer.
Vertical velocity: 3.01 * sin(60) = 3.01 * 0.866025404 = 2.606736465
Time of flight = 2.606736465 * 2 / 9.8 = 0.531987034 seconds.
Horizontal velocity: 3.01 * cos(60) = 3.01 * 0.5 =

8 0

3 years ago