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3 years ago

Please help me ASAP

Mathematics

2 answers:

In-s [12.5K]3 years ago

8 0

Answer:

Step-by-step explanation:

4*4*4=64

vichka [17]3 years ago

6 0

Answer:

Step-by-step explanation:

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What is this answer PLEASE HELP

OLEGan [10]

The answer is D) 5/24 because the fractions have unlike denominators

6 0

2 years ago

Anisha invested $8,000 in an account that earns 10% interest How much money will she have in 15 years of the interest is compoun

tamaranim1 [39]

Answer:

A = $35,198.32

Step-by-step explanation:

Use the formula to calculate compound interest:

A = P(1 + i)ⁿ

"A" for total amount after the time period

"P" for principal, or starting money

"i" for the interest rate in a compounding period

To calculate "i":

i = r / c

"n" for the number of compounding periods

To calculate "n":

n = tc

So, we can combine the formulas into:

$A = P(1+\frac{r}{c})^{tc}$

"c" is the compounding periods in a year. (quarterly = 4)

We know:

P = 8000

r = 10% / 100 = 0.1

t = 15

c = 4

Substitute the information in the formula.

$A = P(1+\frac{r}{c})^{tc}$

$A = 8000(1+\frac{0.1}{4})^{15*4}$ Solve "i" and "n"

$A = 8000(1+0.025)^{60}$ Solve inside the brackets

$A = 8000(1.025)^{60}$ Do the exponent before multiplying by 8000

A = 35198.318 Exact answer

A ≈ 35198.32 Round to two decimal places for money

Therefore she will have $35,198.32 after 15 years.

5 0

3 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.

5 0

3 years ago

Tareq pays $22.10 for 2,6 pounds of salmon, What is the price per pound of the salmon? (20PTS FOR THIS)

Alexeev081 [22]

B

22.10 divided by 2.6 is 8.5 or $8.50

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2 years ago

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AveGali [126]

If you finish the question I may be able to answer it. ^^

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2 years ago

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