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3 years ago

An item on sale cost 85% of the original price of the original price was $40 what is the sales price

Mathematics

1 answer:

irina1246 [14]3 years ago

6 0

Answer:

34$

Step-by-step explanation:

First you take the 85% and convert it into a decimal which is 0.85

Then you multiply 0.85 by 40 and you get 34!

So 85% of 40$ is 34$

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Travka [436]

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A) volume of ingredients

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3 years ago

Read 2 more answers

This glass is 5 in tall and 2.5 inches

Finger [1]

The cubic inches left is 20.05 cubic inches.

<h3 /><h3>Description of the glass </h3>

A glass has the shape of a cylinder. In order to determine which glass is left, the volume of the glass has to be determined. Then what is drank would be subtracted from the volume of the glass.

<h3>Volume of the cylinder. </h3>

Volume of a cylinder = nr^2h

n = 22/7
r = radius= 2.5 / 2 = 1.25

22/7 x 1.25^2 x 5 = 24.55 cubic inches

<h3>Determination of what is left </h3>

24.55 - 4.5 = 20.05 cubic inches

To learn more about to determine the volume of a cylinder, check: brainly.com/question/9624219

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2 years ago

One pizza costs 3 times as much as a soda. If the price of one

Mandarinka [93]

Answer:

Step-by-step explanation:

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3 years ago

a bag contains colored gumdrops 5 are cherry 3 are strawberry 2 are grape. a gumdrop will be randomly selected from the bag what

weqwewe [10]

Answer:

probablity of grapes = 0.2

Step-by-step explanation:

here's the solution : -

probablity of grapes =

$\frac{number \: of \: grapes}{total \: fruits}$

=》

$\frac{2}{2 + 3 + 5}$

=》

$\frac{2}{10}$

=》

$0.2$

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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