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2 years ago

How to solve: a parade traveled 4 miles in 2 hours. how far did tge parade travel per minute?

2 answers:

5 0

Since there is 120 minutes in 2 hours, you would divide 4/120 which would be simplified to 1/30.
Therefore, the parade traveled 1/30 miles per minute or 0.03(a bar notation on 3) miles per minute.

jek_recluse [69]2 years ago

3 0

There's a really easy way to convert any units to other units.

Right now, we have the fraction (4 miles) / (2 hours).

We want to find a fraction that's exactly equal to that one,
but has the units of (miles/minute) or maybe (feet/minute).

Just take the original fraction, and multiply it by some other
fractions.

Each fraction you multiply it by must have the value of ' 1 ' so
you don't change the value of the original fraction. But it can
have different units, that cancel with other units to eventually
give you the units you want.

(4 miles / 2 hours) times (1 hour / 60 minutes)

The second fraction is equal to ' 1 ', because the top and the bottom
have the same value ... 1 hour is the same thing as 60 minutes.

Multiply the fractions: (4 miles x 1 hour) / (2 hour x 60 minutes)

Now you can cancel 'hour' from the top and the bottom, and you have

   (4 miles x 1) / (2 x 60 minutes)

   = (4 miles) / (120 minutes)

   = (4 / 120) mile/minute = 0.0333... mile / minute .

Let's do it again, go a little farther, and get an answer that
might mean more and feel more like an answer.

   (4 miles) / (2 hours) x (5280 feet / mile) x (1 hour / 60 minutes)

The 2nd and 3rd fractions both have the value of ' 1 ', because
the top is equal to the bottom.

Multiply all three fractions:

   (4 miles x 5280 feet x 1 hour) / (2 hours x 1 mile x 60 minutes)

You can cancel both 'mile' and 'hour' out of the top and bottom,
and look what you have left:

   (4 x 5280 feet x 1) / (2 x 1 x 60 minutes)

= (4 x 5280) / (2 x 60) feet / minutes

= (21,120 / 120) feet/minute =    176 feet per minute

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Anton [14]

2000 * .1 = 200 + 2000 = 2200 (First year)

2200 *.1 = 220 + 2200 = 2420 (second year )

2420* .1 = 242 + 2420 = 2662 (third year )

2662 * .1 = 266.2 + 2662 = 2928.2 (fourth year)

They spent $2928.2 on clothing for 4 years.

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3 years ago

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When working with positive and negative numbers, the further left you move on the number line, the________the numbers are and th

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2 years ago

Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that

Maksim231197 [3]

Complete Question

The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?

Answer:

The value is $P(x_1 + x_2 \ge 2 )= 0.5940$

Step-by-step explanation:

From the question we are told that

The rate at which fire breaks out every 10 years is $\lambda = 1$

Generally the probability distribution function for Poisson distribution is mathematically represented as

$P(x) = \frac{\lambda^x}{ k! } * e^{-\lambda}$

Here x represent the number of state which is 2 i.e $x_1 \ \ and \ \ x_2$

Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

$P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 + x_2 \le 1 )$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 = 0)P(x_2 = 0 ) + P( x_1 = 0 ) P( x_2 = 1 )+ P(x_1 = 1 )P(x_2 = 0)$

=> $P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}$

=> $P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]$

$P(x_1 + x_2 \ge 2 )= 0.5940$

3 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

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3 years ago

Solve for x: x^2+4x-5=16x

BigorU [14]

Solve for x: x^2+4x-5=16x
first take 16x from both sides of the equation
x²-12x-5=0
now we have a quadratic. we can solve it by using the quadratic formula
x=-b plus or minus the square root of b²-4ac all divided by 2a
where a=1, b=-12 and c=-5
plug these numbers into the formula
x=12 plus or minus the square root of 144-4x1x-5 all divided by 2
<span>x=12 plus or minus the square root of 164 all divided by 2
or in decimal form </span>x = {12.403124237, -0.403124237}

6 0

2 years ago

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