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3 years ago

9

please help me me with this one question. please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

iren [92.7K]3 years ago

5 0

1st part is u substitute f(x) for 0
0= x+3 over x+1 and then solve which equals -3

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Trisha helped clean up her neighborhood by picking up plastic she collected 3/4 pound of plastic the first day and 1/6 pound of

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She collected more 1/3 pounds on the first day

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3 years ago

Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks

Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀ we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size n = 15 n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean μ₀ = 24,5

Sample mean μ = 25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis H₀ μ = μ₀

Alternative Hypothesis Hₐ μ > μ₀

t (c) = ?

We will define CI = 95 % then α = 5 % α = 0,05 α/2 = 0,025

n = 15 then degree of freedom df = 14

From t-student table we get: t(c) = 2,1448

And t(s)

t(s) = ( μ - μ₀ ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare t(c) and t(s)

t(c) = 2,1448 t(s) = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀ we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

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3 years ago

Find the perimeter of the figure <br> 13.5 ft<br> 9 ft<br> 12.5 ft

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago