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podryga [215]
2 years ago
6

The following graphs show the performance of two stocks over the first five months of the year. A graph titled Stock A B C has m

onth on the x-axis, from 0 to 6, and price on the y-axis, from 0 to 40 in increments of 13.33. A line has points (1, 20), (2, 24), (3, 25), (4, 29), (5, 30). A graph titled Stock X Y Z has month on the x-axis, from 0 to 6, and price on the y-axis, from 20 to 32 in increments of 4. A line has points (1, 20), (2, 24), (3, 25), (4, 29), (5, 30). Which graph appears to show the best performance? Which scale makes the graph appear to be rising more slowly? a. Stock XYZ shows the best performance. The scale of Stock ABC makes the graph appear to be rising more slowly. b. Stock ABC shows the best performance. The scale of Stock ABC makes the graph appear to be rising more slowly. c. Stock XYZ shows the best performance. The scale of Stock XYZ makes the graph appear to be rising more slowly. d. Stock ABC shows the best performance. The scale of Stock XYZ makes the graph appear to be rising more slowly.
Mathematics
2 answers:
hichkok12 [17]2 years ago
8 0

Answer:

C) Stock XYZ shows the best performance. The scale of Stock XYZ makes the graph appear to be rising more slowly.

Step-by-step explanation:

i did this question on my test, from reading the graph, i am almost certain this is the answer. correct me if i am wrong!

BlackZzzverrR [31]2 years ago
6 0

Answer:

B)

Step-by-step explanation:

B because the  Stock XYZ shows the best performance. The scale of Stock ABC makes the graph appear to be rising more slowly.

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What's the correct answer for this?
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\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad  y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
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\bf \textit{\underline{N} is inversely proportional to \underline{A}}\qquad N=\cfrac{k}{A}
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\textit{we also know that }
\begin{cases}
A=3\\
N=16
\end{cases}\implies 16=\cfrac{k}{3}\implies 16\cdot 3=k
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48=k\qquad thus\qquad \boxed{N=\cfrac{48}{A}}\\\\
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4 0
3 years ago
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4 0
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A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 ​min, t
ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

T(t)=T_a+(T_0-T_a)e^{-kt}

where T_a is the ambient temperature, T_0 is the initial temperature, t is the time and k is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say T_0

We know that the ambient temperature is T_a=65, so

T(t)=65+(T_0-65)e^{-kt}

We also know that when t=5 \:min the temperature is T(5)=35 and when t=10 \:min the temperature is T(10)=50 which gives:

T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}

T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}

Rearranging,

35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}

50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}

If we divide these two equations we get

\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}

\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}

Now, that we know the value of k we can use it to find the initial temperature of the beam,

35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5

so the beam started out at 5 °F.

6 0
3 years ago
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