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Sergeeva-Olga [200]
3 years ago
8

The final velocity (v) is given by the formula v = v0 + at, where v0 is initial velocity, v is final velocity, a is acceleration

, and t is time. A car moving at an initial velocity of 20 meters/second accelerates at the rate of 1.5 meters/second2 for 4 seconds. The car’s final velocity is meters/second.
Mathematics
1 answer:
Alona [7]3 years ago
3 0
For this case we have the following equation:
 v = v0 + at
 Where, we have the following values:
 v0 = 20 meters / second
 a = 1.5 meters / second2
 t = 4 seconds
 Substituting values we have:
 v = 20 + (1.5) * 4
 v = 26 meters / second
 Answer:
 
The car's final velocity is:
 
v = 26 meters / second
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Answer:

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Step-by-step explanation:

Given the joint cumulative distribution of X and Y as

F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100

#First find F_x and probability distribution function ,f_x(x):

F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}

#Have determined the probability distribution unction ,f_x(x), we calculate the Expectation of the random variable X:

E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200}  dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\

#We then calculate E(X^2):

E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28

Hence, the Var(X) is 764.28  

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Harlamova29_29 [7]

Answer:

<h2>x = 27°</h2>

Step-by-step explanation:

From the question ( 4x + 7)° and [5(x - 4)]° are vertically opposite

Since vertically opposite angles are equal we can equate them to find x

That's

4x + 7 = 5(x - 4)

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Group like terms

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Hope this helps you

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3 years ago
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