21 divided by 4,857 with remainder

Which equation represents exponential decay?

timama [110]

Answer:

i would say A) But what you gotta do is look up the word of exponential decay and whatever it says will help you with The answer

8 0

3 years ago

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Brittany's family is driving at a speed of 45 miles per hour. If x represents the distance they have traveled and y represents t

Anit [1.1K]

X=45y
x is distance and y is the time the car goes 45 miles per hour so distance equals 45 times time

6 0

3 years ago

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How do I solve this problem? You drive your car on a cold day (20^oF Outside) the engine overheats to about (220^oF). When you p

GaryK [48]

$ln(T- \frac{20}{200})=-0.11t \\ e^{ln(T- \frac{20}{200})}=e^{-0.11t} \\ T- \frac{20}{200} =e^{-0.11t} \\ T-0.1=e^{-0.11t} \\ T=e^{-0.11t}+0.1$
Then, now that we have solved for T, we can evaluate and solve for t=20 minutes.
$T=e^{-0.11t}+0.1 \\ T=e^{-0.11*20}+0.1 \\ T=e^{-2.2}+0.1 \\ T=0.11+0.1 \\ T=0.21$

7 0

3 years ago

How hard must someone pull on the unkown line and inwhat direction to keep the knot from moving

dolphi86 [110]

Answer:

4.4N, 263.4°

Step-by-step explanation:

The sum of all forces must add to zero:

The forces in the x-plane:

∑F = 3N + (cos 120°)*5N + x = 0

x = -(3N + (cos 120°)*5N)

x = -0.5N

The forces in the y-plane:

∑F = (sin 120°)*5N + y = 0

y = -(sin 120°)*5N = -√(3/4)*5N = -4.3N

The magnitude of the unknown force U is:

U = √(x² + y²) = √19 = 4.4N

The direction is given by:

tanФ = y/x

Ф = 263.4°

4 0

3 years ago

A filtration process removes a random proportion of particulates in water to which it is applied. Suppose that a sample of water

Doss [256]

Answer:

$E(Y)=\frac{1}{25}$

Step-by-step explanation:

Let's start defining the random variables for this exercise :

$X_{1}:$ '' The proportion of the particulates that are removed by the first pass ''

$X_{2}:$ '' The proportion of what remains after the first pass that is removed by the second pass ''

$Y:$ '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

$Y=(1-X_{1})(1-X_{2})$

We also assume that $X_{1}$ and $X_{2}$ are independent random variables with common pdf.

The probability density function for both variables is $f(x)=4x^{3}$ for $0$ and $f(x)=0$ otherwise.

The first step to solve this exercise is to find the expected value for $X_{1}$ and $X_{2}$ .

Because the variables have the same pdf we write :

$E(X_{1})= E(X_{2})=E(X)$

Using the pdf to calculate the expected value we write :

$E(X)=\int\limits^a_b {xf(x)} \, dx$

Where $a=$ ∞ and $b=$ - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between $0$ and $1$ ⇒

Using the pdf we calculate the expected value :

$E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}$

⇒ $E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}$

Now we need to use some expected value properties in the expression of $Y$ ⇒

$Y=(1-X_{1})(1-X_{2})$ ⇒

$Y=1-X_{2}-X_{1}+X_{1}X_{2}$

Applying the expected value properties (linearity and expected value of a constant) ⇒

$E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})$

Using that $X_{1}$ and $X_{2}$ have the same expected value $E(X)$ and given that $X_{1}$ and $X_{2}$ are independent random variables we can write $E(X_{1}X_{2})=E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-2E(X)+[E(X)]^{2}$

Using the value of $E(X)$ calculated :

$E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}$

$E(Y)=\frac{1}{25}$

We find that the expected value of the variable $Y$ is $E(Y)=\frac{1}{25}$

3 0

3 years ago