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4 years ago

A color printer prints 21 pages in 10 minutes. How many pages does it print per minute?

1 answer:

5 0

1/
21 pages: 10 minutes
? pages: 1 minute
21÷10=2.1. As a result,it prints 2.1 pages per minute
2/
$3: 5 lbs
?$: 1 lb
3÷5=0.6. As a result, she pays $0.60 per pound of rice. Hope it help!

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3. The three salespeople for a local advertising firm are Lola, Ahmed, and Tommy. Lola sold $2030 in ads, Ahmed sold $1540, and

Salsk061 [2.6K]

3. (a)

Add the three totals.

Total sales for all three: $2030 + $1540 + $1800 = $5370

Now divide each person's sales by the total sales to find each person's fraction of the sales.

Lola: 2030/5370
Ahmed: 1540/5370
Tommy: 1800/5370

3. (b)

Multiply each fraction from part (a) by $100.

Lola: 2030/5370 * $100 = $37.80
Ahmed: 1540/5370 = $28.67
Tommy: 1800/5370 = $33.52

6 0

3 years ago

Marie's journal is 400 pages long. She has used 20% of the journal. How many pages has she used so far?

vitfil [10]

Answer:

40/ or /80

Step-by-step explanation:

Because 400 is a lot and if it's 100% is 400 right so we do 80% what would that be? we don't know right.

What about 100 pages that would be (im pretty sure) 40% or 50%.

I hope this helped!!!

GOODLUCK!!!!!!!!!!

5 0

3 years ago

Read 2 more answers

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

Sherman has 3 cats and 2 dogs. he wants to buy a toy for each of his pets. Sherman has 22$ to spend on pet toys. How much can he

WARRIOR [948]

Sherman wants to spend on each pet, so we are assuming he wants to spend the same amount. He has 5 pets in all (3 cats, 2 dogs)

Divide 22 with 5

22/5 = 4.4

He can spend $4.40, or in fraction form:

4 4/10 (Mixed fraction)

hope this helps

6 0

3 years ago

Question 15 of 26

DanielleElmas [232]

Answer:

3.5 yards

Step-by-step explanation:

3 0

3 years ago