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OleMash [197]
3 years ago
11

Ellianna made a scatter plot that showed a positive association and had one outlier. Write a possible scenario for Ellianna's sc

atter plot
Hint
o
The scatter plot could show test scores based on number of student absences. The outlier could be a student with high test scores who has many
absences,
The scatter plot could show the amount of pay workers at a factory received based on the number of years they have worked at the factory. The
outlier could be someone who is very skilled and has learned her job quickly.
The scatter plot could show the amount of weight that someone gains based on the number of calories that they consume. The outlier could be
someone who eats few calories and does not gain much weight.
The scatter plot could show the amount of weight that a elephant gains based on its age. The outlier could be an adult elephant that weighs more
than a baby elephant.
Mathematics
1 answer:
zzz [600]3 years ago
5 0

Answer:

not sure ask teacher for help

Step-by-step explanation:

ok

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5 yes or good i'm not good morning guys

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What is the equation of the following line written in slope-intercept form? the x intercept of 2.
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Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)
zloy xaker [14]

Answer:

A)

k=0

B)

\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}

C)

\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}

Step-by-step explanation:

We are given the function:

\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}

A)

Given that h(1) = 20, we want to find <em>k</em>.

h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

\displaystyle (20) = 20e^{k(1)}

Simplify:

1= e^k

Anything raised to zero (except for zero) is one. Therefore:

k=0

B)

Given that h(1) = 40, we want to find 2<em>k</em> + 1.

Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

\displaystyle (40) = 20e^{k(1)}

Simplify:

\displaystyle 2 = e^{k}

We can take the natural log of both sides:

\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}

By definition, ln(e) = 1. Hence:

\displaystyle k = \ln 2

Therefore:

2k+1 = 2\ln 2+ 1 \approx 2.3863

C)

Given that h(1) = 10, we want to find <em>k</em> - 3.

Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

\displaystyle (10) = 20e^{k(1)}

Simplfy:

\displaystyle \frac{1}{2} = e^k

Take the natural log of both sides:

\displaystyle \ln \frac{1}{2} = k\ln e

Therefore:

\displaystyle k = \ln \frac{1}{2}

Therefore:

\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931

3 0
3 years ago
A tank contains 250 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
VashaNatasha [74]

Answer:

A(t)=250-210e^{-\frac{t}{50}}

Step-by-step explanation:

We are given that

Volume,V=250 L

Mass of salt=m=40 g

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Pumped into the tank at the rate=5 L/min

We have to find the number A(t) of grams of salt in the tank at time t.

Rate of change of salt in the tank

\frac{dA}{dt}=Rate in-Rate out

Rate in=5 L/min

Rate out=\frac{A}{250}\times 5=\frac{A}{50} L/min

\frac{dA}{dt}=5-\frac{A}{50}=\frac{250-A}{50}

\int \frac{50dA}{250-A}=\int dt

-50ln(250-A)=t+C

Using the formula

\int \frac{dx}{x}=ln x

A=40 and t=0

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-50ln(210)=C

Substitute the value

-50ln(250-A)=t-50ln(210)

-50ln(250-A)+50ln(210)=t

50ln\frac{210}{250-A}=t

t=50ln\frac{210}{250-A}

\frac{t}{50}=ln\frac{210}{250-A}

\frac{210}{250-A}=e^{\frac{t}{50}

\frac{250-A}{210}=e^{-\frac{t}{50}}

250-A=210e^{-\frac{t}{50}}

A(t)=250-210e^{-\frac{t}{50}}

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