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2 years ago

A student has 6 nickel 14 dimes and 8 quarter. What is the total number of coins?

Mathematics

1 answer:

S_A_V [24]2 years ago

7 0

There is 28 coins total

You might be interested in

Find the sum of X + y, when x= 3.1, and y= 5.4

Mashutka [201]

So all this is is changing the x and y variables,

In this case, x = 3.1 and y = 5.4 so,

3.1 + 5.4

3 + 5 = 8

.1 + .4 = .5

8.5

4 0

3 years ago

Learning Theory In a typing class,the averege number N of words per minutes typed after t weeks of lessons can be modeled by N =

Jet001 [13]

Answer:

a) $N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

b) $N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

c) $70 =\frac{95}{1+8.5 e^{-0.12t}}$

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

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

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

Step-by-step explanation:

For this case we have the following expression for the average number of words per minutes typed adter t weeks:

$N(t) = \frac{95}{1+8.5 e^{-0.12t}}$

Part a

For this case we just need to replace the value of t=10 in order to see what we got:

$N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

So the number of words per minute typed after 10 weeks are approximately 27.

Part b

For this case we just need to replace the value of t=20 in order to see what we got:

$N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

So the number of words per minute typed after 20 weeks are approximately 54.

Part c

For this case we want to solve the following equation:

$70 =\frac{95}{1+8.5 e^{-0.12t}}$

And we can rewrite this expression like this:

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

Now we can divide both sides by 8.5 and we got:

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

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

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

And then if we solve for t we got:

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

And we can see this on the plot 1 attached.

Part d

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

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

8 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccup

Mkey [24]

The given sample's 95% confidence interval is 11.6 ± 0.5357. I.e., from the lower bound 11.06 to the upper bound 12.14. Its margin error is 0.5357.

<h3>How to find the confidence interval?</h3>

To find the confidence interval C.I,

Determine the mean(μ) of the sample
Determine the standard deviation(σ)
Determine the z-score for the given confidence level using the z-score table
Using all these, the confidence interval is calculated by the formula, C. I = μ ± z(σ/√n) where n is the sample size and z(σ/√n) gives the margin error. So, we can also write C. I = mean ± margin error.

<h3>Calculation:</h3>

It is given that,

Sample size n = 225

Sample mean μ = 11.6

Standard deviation σ = 4.1

Since 95% confidence interval, z-score is 1.96

So, the required confidence interval is calculated by,

C. I = μ ± z(σ/√n) or mean ± margin error

On substituting,

C. I = 11.6 ± 1.96(4.1/√225)

= 11.6 ± 1.96 × 0.2733

= 11.6 ± 0.5357

So, the lower bound = 11.6 - 0.5357 = 11.065 and

the upper bound = 11.6 + 0.5357 = 12.135.

Thus, the confidence interval is from 11.06 to 12.14, and its margin error is 0.0537 for the population mean number of unoccupied seats per flight.

Learn more about confidence intervals here:

brainly.com/question/16236451

#SPJ4

3 0

2 years ago

What are the roots of the equation?<br><br> x^2(6x−11)(7x−8)=0

trasher [3.6K]

multiply all the factors in x^2

to get the roots which are 0, 11/6 and 8/7

3 0

3 years ago

Read 2 more answers

Yo help me with number 6 pls thx and tell me how. I can't find the answer

dybincka [34]

First find the formula which is A(N)=A+(N-1)D then fill in what you know
N=term your looking for
A=first term
D= common diffrence

Once filled in you will get this 18(18)=18+(18-1)-6
then solve to get 324= -84

Your final answer is that the 18th term is -84

Hope I could help

6 0

3 years ago