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3 years ago

Henry can write 5 pages of his novel in 3 hours. at this rate, how many pages can henry write in 888 hours?

2 answers:

3 0

If he writes 5 pages in 3 hours, you take 888 divide it by 3 and get 296. Then, you write 5 and 3 as a fraction, 5/3. Multiply both by 296 and you get 1480 pages in 888 hours.

I hope this helps!

lesya692 [45]3 years ago

3 0

Write and solve an equation of ratios:

5 pages p
------------- = ------------- Cross multiply. Solve for p.
3 hrs 888 hrs

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GETS BRAINLIEST AND 100 POINTS!!!!! FIND THE INTERVAL OF THE CONVERGENCE

harina [27]

Answer:

Look at explanation

Step-by-step explanation:

Ok,

I got...

radius of convergence = 1

interval of convergence = 3 ≤ x ≤ 5

8 0

3 years ago

Rewrite the equation of the line 5y = -1/4x - 6 in standard form

netineya [11]

Answer:

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C

Given

5y = - $\frac{1}{4}$ x - 6 ( multiply through by 4 to clear the fraction )

20y = - x - 24 ( add x to both sides )

x + 20y = - 24 → D

7 0

3 years ago

In 2015, Oklahoma experienced 907 perceptible earthquakes (far surpassing California), for an average of about 2.5 perceptible e

Rufina [12.5K]

Answer:

The probability P of a day with no perceptible earthquakes is 0.0821.

Step-by-step explanation:

We will consider that earthquakes occurring in a day is a Poisson process. The following Poisson probability distribution formula will be used in this question.

p(x,λ) = [e^-λ (λ)ˣ]/x!

where x = number of outcomes occurring

λ = mean number of occurrences

(a) So, in this question we have λ = 2.5 and we need to find the probability that x=0 (no perceptible earthquakes in a day). So,

P(X=0) = p(0,2.5) = [(e^-2.5)(2.5)⁰]/0!

= ((0.0821)*1)/1

P(X=0) = 0.0821

The probability P of a day with no perceptible earthquakes is 0.0821.

6 0

4 years ago

A 500 gallon tank initially contains 200 gallons of water with 5 lbs of salt dissolved in it. Water enters the tank at a rate of

Lapatulllka [165]

Until the concerns I raised in the comments are resolved, you can still set up the differential equation that gives the amount of salt within the tank over time. Call it $A(t)$ .

Then the ODE representing the change in the amount of salt over time is

$\dfrac{\mathrm dA}{\mathrm dt}=\text{rate in}-\text{rate out}$
$\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{\frac15(1+\cos t)\text{ lbs}}{1\text{ gal}}-\dfrac{2\text{ gal}}{1\text{ hr}}\times\dfrac{A(t)\text{ lbs}}{500+(2-2)t}$
$\dfrac{\mathrm dA}{\mathrm dt}=\dfrac25(1+\cos t)-\dfrac1{250}A(t)$

and this with the initial condition $A(0)=5$

You have

$\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}A(t)=\dfrac25(1+\cos t)$
$e^{t/250}\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{250}e^{t/250}A(t)=\dfrac25e^{t/250}(1+\cos t)$
$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/250}A(t)\right]=\dfrac25e^{t/250}(1+\cos t)$

Integrating both sides gives

$e^{t/250}A(t)=100e^{t/250}\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+C$
$A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)+Ce^{-t/250}$

Since $A(0)=5$ , you get

$5=100\left(1+\dfrac1{62501}\right)+C\implies C=-\dfrac{5937695}{62501}$

so the amount of salt at any given time in the tank is

$A(t)=100\left(1+\dfrac1{62501}\cos t+\dfrac{250}{62501}\sin t\right)-\dfrac{5937695}{62501}e^{-t/250}$

The tank will never overflow, since the same amount of solution flows into the tank as it does out of the tank, so with the given conditions it's not possible to answer the question.

However, you can make some observations about end behavior. As $t\to\infty$ , the exponential term vanishes and the amount of salt in the tank will oscillate between a maximum of about 100.4 lbs and a minimum of 99.6 lbs.

5 0

4 years ago

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

$S_{1893} =5632.98$

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

$S_{1893} =5632.98$ ----- approximated

4 0

3 years ago