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3 years ago

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A car's rear windshield wiper rotates 130°. The total length of the wiper mechanism is 25 inches and the length of the wiper bla

de is 14 inches. Find the area wiped by the wiper blade. (Round your answer to one decimal place.)

1 answer:

Delicious77 [7]3 years ago

8 0

The area wiped by the viper blade is approximately 486.7 square inches.

The area wiped by the wiper blade ( $A$ ), in square inches, is equal to the area of the circular section described below:

$A = \frac{\theta\cdot \pi}{360}\cdot (R^{2}-r^{2})$ (1)

Where:

$\theta$ - Angle, in sexagesimal degrees.
$r$ - Inner radius, in inches.
$R$ - Outer radius, in inches.

If we know that $\theta = 130^{\circ}$ , $R = 25\,in$ and $r = 14\,in$ , then the area wiped by the wiper blade is:

$A = \frac{130\pi}{360} \cdot [(25\,in)^{2}-(14\,in)^{2}]$

$A \approx 486.7\,in^{2}$

The area wiped by the viper blade is approximately 486.7 square inches.

We kindly invite to check this question on circular sections: brainly.com/question/14989543

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The number of people arriving for treatment at an emergency room can be modeled by a Poisson Distribution with a rate parameter

AlekseyPX

Answer:

a) 0.052 = 5.2% probability that exactly three arrivals occur during a particular hour

b) 0.971 = 97.1% probability that at least three people arrive during a particular hour

c) 5.25 people are expected to arrive during a 45-min period

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson Distribution with a rate parameter of seven per hour.

This means that $\mu = 7$

(a) What is the probability that exactly three arrivals occur during a particular hour?

This is P(X = 3).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.052$

0.052 = 5.2% probability that exactly three arrivals occur during a particular hour.

(b) What is the probability that at least three people arrive during a particular hour? (Round your answer to three decimal places.)

Either less than three people arrive, or at least three does. The sum of the probabilities of these events is 1. So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$ , which is

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.001$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.006$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.022$

So

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.001 + 0.006 + 0.022 + 0.029$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.029 = 0.971$

0.971 = 97.1% probability that at least three people arrive during a particular hour

(c) How many people do you expect to arrive during a 45-min period?

During an hour(60 minutes), 7 people are expected to arrive. So, using proportions:

$\frac{45*7}{60} = \frac{3*7}{4} = \frac{21}{4} = 5.25$

5.25 people are expected to arrive during a 45-min period

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3 years ago

Both copy machines reduce the dimensions of images that are run through the machines. Which statement is true about the results

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off a

zubka84 [21]

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a = (v - u)/t = (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' = 17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

4 0

2 years ago

Francine ran 3 1/2 laps around the circular track in the school gym. The track has a diameter of 1/2 miles. About how far does F

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Answer:

Francine run 5.495 miles

Step-by-step explanation:

Given :

Francine ran $3\frac{1}{2}$ laps around the circular track in the school gym.

The track has a diameter of 1/2 miles.

To Find : About how far does Francine run?

Solution :

First we need to calculate the perimeter of circular track :

Perimeter of circle : $2\pi r$

Since we are given the diameter = 1/2 miles = 0.5 miles

So, Radius will be (Diameter/2) = 0.5/2 = 0.25 miles

So, The perimeter will be:

$2\pi *0.25$

$0.5\pi$

Since $\pi =3.14$

⇒ $0.5*3.14$

⇒ $1.57$

Thus the circular track is of 1.57 miles .

Since she ran $3\frac{1}{2}$ = 7/2=3.5 laps

1 lap = 1.57 miles

So, 3.5 laps = 1.57*3.5 =5.495 miles .

Hence , Francine run 5.495 miles

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3 years ago