All categories

2 years ago

4n + (n - 5) +2 dasdasfasfafasf

Mathematics

2 answers:

gogolik [260]2 years ago

8 0

The answer i got is n = -13

suter [353]2 years ago

7 0

Answer:

5n-3

Step-by-step explanation:

4n + (n - 5) +2

combine like terms, 4n and n

5n + (-5) +2

combine like terms again, -5 and 2

5n-3

You might be interested in

A rectangular prism is 15 centimeters long, 33 centimeters wide, and 29 centimeters high. What is the surface area of the rectan

Mama L [17]

Answer:

add 15 + 33+29 =77

Step-by-step explanation:

mark me brainliest

4 0

1 year ago

Read 2 more answers

Which decimal is equivalent to<br> 4/20

Anastaziya [24]

The answer should be .20

6 0

2 years ago

The notation __1____ reads the probability of Event B given that Event A has occurred. If Events A and B are independent, then t

telo118 [61]

P(B|A) (option B)

Doesn't affect (option A)

P(B|A) = P(B) (option A)

Explanation:

1) Conditional probabilities could be in the form P(A|B) or P(B|A)

P(B|A) is a notation that reads the probability of event B given that event A has occurred.

P(B|A) (option B)

2) Independent events do not affect the outcome of each other

For event A and B to be independent, the probability of event A occurring doesn't affect the the probability of event B occurring

Doesn't affect (option A)

3) Events A and B are independent if the following are satisfied:

P(A|B) = P(A)

P(B|A) = P(B)

The ones that appeared in the option is P(B|A) = P(B) (option A)

3 0

8 months 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

2 years ago

You have $16 and a coupon for a $5 discount at a local

Brrunno [24]

8 bags of chips total

7 0

2 years ago