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Svetradugi [14.3K]
2 years ago
5

An object moves 100 m in 4 s and then remains at rest for an additional 1 s.

Mathematics
1 answer:
nydimaria [60]2 years ago
8 0

Answer:

20 m per second

Step-by-step explanation:

100 / 5 = 20

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Determine if the equation y = 2/5 x − 4 y=25x-4 represents a proportional relationship. If so, determine the constant of proport
Vilka [71]

Answer:

The linear equation represents a proportional relationship and its constant of proportionality is k = \frac{2}{5}.

Step-by-step explanation:

A proportional relationship exists when the following relationship is observed:

u = k\cdot v

Where:

u - Dependent variable.

v - Independent variable.

k - Proportionality constant.

If y =\frac{2}{5}\cdot x - 4 and v = x and u = y+4, the following expresion is found:

y = \frac{2}{5}\cdot x -4

y + 4 = \frac{2}{5}\cdot x

u = \frac{2}{5}\cdot v

The linear equation represents a proportional relationship and its constant of proportionality is k = \frac{2}{5}.

8 0
3 years ago
Nicole works 50 hours a week and earns $25 per hour. Her company pays overtime of one and half times for all hours worked over 4
ankoles [38]
Her gross weekly pay would be $1375.

25 * 40 = $1000

time and a half means she would make $37.50 for the 10 hours overtime she works. 10 * 37.5 = 375

375 + 1000 = $1375
4 0
3 years ago
If TR=11 what is the length of RS.
masha68 [24]
No se porque no se me van a hacer nada de eso
6 0
2 years ago
Geraldo is trying to fit his encyclopedias on the shelf. Each encyclopedia is 2 1/4 inches thick. If Geraldo shelf is 2 1/4 feet
Maurinko [17]

Answer:

12

Step-by-step explanation:

First, find out how wide his shelf is in inches. 2.25*12=27. So his shelf is 27 inches wide. Then divide that by 2.25 to find the number of encyclopedias will fit.

8 0
3 years ago
The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand
Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})

P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})

P(X > 1560) = P(Z > \dfrac{110}{220})

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean \mu_x = np = 52 × 0.3085 = 16.042

The standard deviation =  \sqrt {n \time p (1-p)}

The standard deviation = \sqrt {52 \times 0.3085 (1-0.3085)}

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})

Pr ( Y > 20) = P(Z >1 .338)

From z tables

P(Y > 20) \simeq 0.0903

7 0
3 years ago
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