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Georgia [21]
3 years ago
5

The 32 students in Ms.Byrd's art class made a project with construction paper. Eleven students used red construction paper, 8 st

udents used green construction paper, and the rest of the students used blue construction paper. If Ms. Byrd chooses one project at random, what is the probability it will be blue? Plz explain and show work.
A. 1/4
B. 11/32
C. 13/32
D.2/3
Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer: The correct answer is C. 13/32

Step-by-step explanation: There are 32 students, and 11 + 8 = 19. SO in order to figure out how many students used blue, we need to subtract 19 from 32. that gives us 13. So:

13 out of 32 students, or

13/32

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Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

=========================================================

Explanation:

Let f(n) = \frac{(-1)^{n+1}}{n!}

The summation given to us represents the following

\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\

There are infinitely many terms to be added.

-------------------

The partial sums only care about adding a finite amount of terms.

The partial sum S_1 is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say S_1 = f(1) = 1

I'm skipping the steps to compute f(1) since you already have done so.

-------------------

The second partial sum is when things get a bit more interesting.

We add the first two terms.

S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\

The scratch work for computing f(2) is shown in the diagram below.

-------------------

We do the same type of steps for the third partial sum.

S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\

The scratch work for computing f(3) is shown in the diagram below.

-------------------

Now add the first four terms to get the fourth partial sum.

S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like S_3 = S_2 + f(3) and S_4 = S_3 + f(4)

In general, S_{n+1} = S_{n} + f(n+1) so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

-------------------

Let's use that recursive trick to find S_5

S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333

The scratch work for f(5) is shown below.

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Answer:

  -25            

 ——— = -2.08333

    12  

Step-by-step explanation:

Step  1  :

           7

Simplify   —

           4

Equation at the end of step  1  :

       1           7

 (0 -  —) +  (0 -  —)

       3           4

Step  2  :

           1

Simplify   —

           3

Equation at the end of step  2  :

       1     -7

 (0 -  —) +  ——

       3     4

Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

     The left denominator is :       3

     The right denominator is :       4

       Number of times each prime factor

       appears in the factorization of:

Prime

Factor   Left

Denominator   Right

Denominator   L.C.M = Max

{Left,Right}

3 1 0 1

2 0 2 2

Product of all

Prime Factors  3 4 12

     Least Common Multiple:

     12

Calculating Multipliers :

3.2    Calculate multipliers for the two fractions

   Denote the Least Common Multiple by  L.C.M

   Denote the Left Multiplier by  Left_M

   Denote the Right Multiplier by  Right_M

   Denote the Left Deniminator by  L_Deno

   Denote the Right Multiplier by  R_Deno

  Left_M = L.C.M / L_Deno = 4

  Right_M = L.C.M / R_Deno = 3

Making Equivalent Fractions :

3.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

  L. Mult. • L. Num.      -1 • 4

  ——————————————————  =   ——————

        L.C.M               12  

  R. Mult. • R. Num.      -7 • 3

  ——————————————————  =   ——————

        L.C.M               12  

Adding fractions that have a common denominator :

3.4       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

-1 • 4 + -7 • 3     -25

———————————————  =  ———

      12            12

Final result :

 -25            

 ——— = -2.08333

 12            

Processing ends successfully

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