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maksim [4K]
3 years ago
11

What is y=3(2x+1)+5 work need to be shown

Mathematics
1 answer:
jonny [76]3 years ago
6 0

Answer:

In slope intercept form, this equation is y = 6x + 8

Step-by-step explanation:

All you need to do to get this to slope intercept form is to distribute the term and then combine like terms.

y = 3(2x + 1) + 5 ----> Distribute

y = 6x + 3 + 5 ----> Combine like terms

y = 6x + 8

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Suppose that $9500 is placed in an account that pays 8% interest compounded each year.
Blababa [14]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$9500\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{each year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}


\bf A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 1}\implies A=9500(1.08)\implies \boxed{A=10260} \\\\\\ \stackrel{\textit{after 2 years, t = 2}}{A=9500\left(1+\frac{0.08}{1}\right)^{1\cdot 2}}\implies A=9500(1.08)^2\implies \boxed{A=11080.8}

6 0
3 years ago
What is a mathematical statement that two expressions are equal
natulia [17]
It is called an equation.
6 0
2 years ago
Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1
elena-s [515]

Answer:

a)P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

b) P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

c) 0.75 =\frac{0.83}{1+e^{-0.2n}}

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

d) If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

P(n) = \frac{0.83}{1+e^{-0.2t}}

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

So the number of correct reponses  after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

0.75 =\frac{0.83}{1+e^{-0.2n}}

And we can rewrite this expression like this:

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

Now we can apply natural log on both sides and we got:

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

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egoroff_w [7]

Answer:

look at explanation

Step-by-step explanation:

a) -13,-14,-15,-16

b) -2,-3,-4,-5

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Margie walked 3.75 miles on Tuesday morning. On Wednesday morning, she walked 1.2 times as far as she did on Tuesday. How far di
kari74 [83]
She walked 4.5 miles. just multiply the miles she walked Tuesday by 1.2.
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2 years ago
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