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Tasya [4]
3 years ago
9

Learning math. The Core Plus Mathematics Project (CPMP) is an innovative approach to teaching mathematics that engages students

in group investigations and mathematical modeling. After field tests in 36 high schools over a three-year period, researchers compared the perfor­mances of CPMP students with those taught using a traditional curriculum. In one test, students had to solve applied algebra problems using calculators. Scores for 320 CPMP students were compared with those of a control group of 273 students in a traditional math program. Computer software was used to create a confidence interval for the difference in mean scores (Journal for Research in Mathematics Education, 31, no. 3, 2000).
Conf. level: 95%

Variable: µ(CPMP) — µ(Ctrl)

Interval: (5.573, 11.427)

a) What is the margin of error for this confidence interval?

b) If we had created a 98% confidence interval, would the margin of error be larger or smaller?

c) Explain what the calculated interval means in this con­text.

d) Does this result suggest that students who learn mathe­matics with CPMP will have significantly higher mean scores in applied algebra than those in traditional programs? Explain.
Mathematics
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

D is the answer

Step-by-step explanation:

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Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1
mafiozo [28]

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $ b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value x_{0}  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]

Applying chain rule,

y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

7 0
4 years ago
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