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Leni [432]
2 years ago
8

Say the distance traveled is 7.2 meters and the time it took to get there was .28 seconds. To find the average speed, you would

have to divide the distance ( 7.2 meters ) by the time (.28 seconds ) which is around 25.7 meters per second. How can it be 25.7 meters per second if the total distance traveled was 7.2 meters? The meters in average speed is greater than the distance? Shouldn’t the meters per second be way smaller than the actual distance? Why is it greater? Could someone please help me understand this?
Mathematics
1 answer:
lara [203]2 years ago
5 0

fomular you say :speed =distance taveled divide by time traveled .......hope this will help

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Ilu<br> y = x2 – 16x + 63?
Neporo4naja [7]

Answer:

y=-14x+63

Step-by-step explanation:

this is the most simplified equation

7 0
2 years ago
The speed at which cars travel on the highway has a normal distribution with a mean of 60 km/h and a standard deviation of 5 km/
oee [108]

The z-score of the speed value gives the measure of dispersion of the from

the mean observed speed.

The probability that the speed of a car is between 63 km/h and 75 km/h is

<u>0.273</u>.

The given parameters are;

The mean of the speed of cars on the highway, \overline x = 60 km/h

The standard deviation of the cars on the highway, σ = 5 km/h

Required:

The probability that the speed of a car is between 63 km/h and 75 km/h

Solution;

The z-score for a speed of 63 km/h is given as follows;

Z=\dfrac{x-\bar x }{\sigma }

Which gives;

Z=\dfrac{63-60 }{5 } = 0.6

From the z-score table, we have;

P(x < 63) = 0.7257

The z-score for a speed of 75 km/h is given as follows;

Z=\dfrac{75-60 }{5 } = 3

Which gives, P(x < 75) = 0.9987

The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;

P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273

The probability that the speed of a car is between 63 km/h and 75 km/h is

<u>0.273</u>.

Learn more here:

brainly.com/question/17489087

7 0
2 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui
Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= \frac{dx}{dt}=1. 8 m/s

\triangle ABD\sim\triangle ECD

Therefore, \frac{AB}{EC}=\frac{BD}{CD}

\frac{y}{2}=\frac{12}{x}

xy=24

Differentiate w.r.t t

x\frac{dy}{dt}+y\frac{dx}{dt}=0

x\frac{dy}{dt}=-y\frac{dx}{dt}

\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}

When the man is 4 m from  the building

Then, we have x=12-4=8 m

\frac{dx}{dt}=1.8 m/s

Substitute the values in above equation then, we get

8y=24

y=\frac{24}{8}=3

Substitute the values then we get

\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0
3 years ago
Which of the following points are solutions to the system of inequalities
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2 years ago
harvey collects rare marbles. He has 125 marbles in collection 1, 36 fewer marbles in collection 2, and 53 more marbles than col
Usimov [2.4K]

Collection 1, no of marbles = 125

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Hence total marbles = 125+89+178

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Thus division is used for finding out the no of marbles in a tray.

For finding out no of marbles in 3 trays, we get = 3x27 = 81 marbles

(Here direct variation is used)

Answer is 81 marbles

6 0
3 years ago
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