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prohojiy [21]
3 years ago
12

Mrs. diaz makes 5 dozen cookies for her class. one ninth of her 27 students are absent the day she brings the cookies. if she sh

ares the cookies equally among the students who are present, how many cookies will each student get
Mathematics
2 answers:
Savatey [412]3 years ago
8 0
We know that Mrs. Diaz has a total of 60 cookies (which is 5 * 12 = 60).

To find how many students were absent, we divide the 27 by 1/9

To do this, we set the 27 into a fraction, which would be 27/1

Next we would flip the other fraction to be 1/27

Now we multiply.

27 * 1 and 9 * 1

27 / 9

When we simplify it, we get 3, which means 3 students were absent that day from Mrs. Diaz's class.

So, knowing that there are sixty cookies and 24 students, we divide.

60 / 24 = 2.5

Meaning that each student would get two cookies each, giving Mrs. Diaz 12 leftovers.

Hope this helped! c:

- Kat
saw5 [17]3 years ago
6 0
Each student gets 2.5 cookies.
We know that there are 24 students there and 60 cookies, (since one ninth of 27 is 3 so 3 students are not there, and 24 are, and that 5 dozen is another way to write 60) we need to divide 60 by 24 which is 2.5.
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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
Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown.
shepuryov [24]

The smallest possible whole-number length of the unknown side is 17 inches.

<h3>What is the Pythagoras theorem?</h3>

The Pythagoras theorem states that the square of the longest side must be equal to the sum of the square of the other two sides in a right-angle triangle.

From the information given, the sides of an obtuse triangle measure 9 inches and 14 inches.

Therefore, the third side will be:

c² = 9² + 14²

c² = 81 + 196

c² = 277

c = ✓277

c = 16.64

c = 17

Hence, the smallest possible whole-number length of the unknown side is 17 inches.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ1

7 0
1 year ago
Complete the inequality statement with the symbol that makes it true.
OLga [1]
Equal because these numbers are of the same value
7 0
3 years ago
What's the perimeter of a rectangle with length 12 m and width 5 m
Pie
12+12=24+5+5=34 meters squared
4 0
3 years ago
Need the answer right now
hodyreva [135]

Answer:

Diameter is 19

Step-by-step explanation:

6 0
3 years ago
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