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marta [7]
3 years ago
9

A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the poli

ce car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road
Mathematics
1 answer:
pishuonlain [190]3 years ago
6 0

Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

Answer:

The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.  

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

h² = 19400

h = √19400

h = 139.284 ft

Again;

Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt  be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

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3 years ago
A small junior college had an enrollment of 7,843 students in the year 1990 and 8,793 students in the year 2000. Let x=0 represe
abruzzese [7]

Correct answer is: (0,7843) and (10,8793)

Solution:-

Given that a junior college has an enrollment of 7843 students in 1990 and 8793 students in year 2000.

We have to write this data as (x,y) .

Where x= years after 1990 and y=number of students enrolled.

Since in 1990, 7843 students enrolled, x = 1990-1990=0

And y=7843.

Hence one ordered pair is (0,7843).

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4 0
3 years ago
I need help on this question
Doss [256]
Hi,
Following the rules of PEMDAS you first subtract in the parentheses. 
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4 0
3 years ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
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