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3 years ago

Now see if you can solve these word problems about time and distance.

Mathematics

1 answer:

ladessa [460]3 years ago

4 0

Answer:

3.5 hours

Step-by-step explanation:

Ediburgh -------------------------------350 miles-------------------------London

Let the distance traveled by train from Edin to London be "x"

hence the distance traveled by train from London to Edin would be "350-x"

The rate of Edin to London train is 60
The rate of London to Edin train is 40

Let the time they meet be"t"

Now, we know distance formula to be D = RT

Where

D is distance

R is rate

and

T is time

<u>Train from Edin to London:</u>

x = 60t

<u>Train from London to Edin:</u>

350-x = 40t

Solving for x:

350 - 40t = x

We put this into first equation:

x = 60t

350 - 40t = 60t

350 = 100t

t = 350/100

t = 3.5

They meet after 3.5 hours

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konstantin123 [22]

For the equation F(x) = ax² + bx + c we have:

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F(x) = -3x² + 18x + 3 ⇒ a = -3, b = 18

a < 0 ⇒ the function has a maximum value

Quadratic function has the maximum value (or minimum) at vertex of its parabola.

The maximum value is k at x=h where: $h=\dfrac{-b}{2a}$ and k = F(h)

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Therefore:

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1 year ago

F(x)=5(x-2) <br> find x if f(x)=-20

Lapatulllka [165]

Answer:

x = -2

Step-by-step explanation:

f(x)= 5(x-2)

since, f(x)= -20

-20 = 5(x-2)

x-2 = -20/5

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x = -2

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2 years ago

Find the equation of the line passing through the point (1, 3) and parallel to the line y--5x 2 Select one: ? ?.ys x+14 3 O E. n

Varvara68 [4.7K]

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Step-by-step explanation:

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3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

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