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Shtirlitz [24]
2 years ago
13

Cassie made a picture graph to record the points that third grade students scored on test.Mrs Wilson's group scored 40 points in

all.There are 11 students in Mrs Wilson's group.What is one way Mrs Wilson's group may have scored 40 points.Tell how you can be precise when solving this problem.And also so a math words and symbol for solving this problem
Mathematics
1 answer:
lara31 [8.8K]2 years ago
6 0

Answer:

4 students scored 3 points each

7 students scored 4 points each

Step-by-step explanation:

<u>Undetermined equations </u>

There are problems where there are not enough data to provide a unique (determined) solution. We form as many equations as possible and try to find the combination of values of the variables to fulfill all the conditions

For this question, we are assuming the grades go from 1 to 5. We know 11 students in Mrs. Wilson's group scored 40 points in all. We must find a combination of points to produce a sum of 40. We're also assuming only integer points can be scored.

To make things easier, let's find the mean value of the total scores: 40/11=3.6

If we stick to the closest integer near 3.6 we can find a quick solution. Let's think all the students scored 4. It would give a total of 44 points (4 more than required). To adjust the excess, we simply set the score of 4 students to 3 points. Our solution will be

4 students scored 3 points each: 12 points

7 students scored 4 points each: 28 points

Total: 40 points

Note: You can find more solutions apart from this one. For example, set the score of one of the 4-points students to 3. To compensate, set another one from that group to 5 points. Our new solution will be

5 students scored 3 points each: 15 points

5 students scored 4 points each: 20 points

1 student  scored 5 points     :  5 points

It also has a total of 40 points from 11 students

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2 years ago
The winning long jump at a track meet was 27 ft 10 in. Convert this distance to meters. Round to the nearest hundredth.
Allushta [10]

1 inch = 2.54 cm

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6 0
1 year ago
Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
musickatia [10]
(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}
\implies r^4-5r^3+6r^2+4r-8=0
\implies (r-2)^3(r+1)=0\implies r=2,r=-1

So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n

(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n
G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)
G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)
(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3
G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
3 0
2 years ago
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