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Oksana_A [137]
3 years ago
6

Using traditional methods it takes 101 hours to receive an advanced driving license. A new training technique using Computer Aid

ed Instruction (CAI) has been proposed. A researcher believes the new technique may reduce training time and decides to perform a hypothesis test. After performing the test on 280 students, the researcher decides to reject the null hypothesis at a 0.05 level of significance.
What is the conclusion?
a. There is sufficient evidence at the 0.02level of significance that the new technique reduces training time.
b. There is not sufficient evidence at the 0.02 level of significance that the new technique reduces training time.
Mathematics
1 answer:
babunello [35]3 years ago
6 0
The answer will be B
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Question 16 of 22What is the product of (2x+3x - 1) and (3x+5)?A. 6x3 +19x2 +12x-5B. 6x3 + 10x2 +15x-5C. 6x3 +9x? - 3x-5D. 6x3 +
In-s [12.5K]

Given: (2x+3x - 1) and (3x+5)

The product of them will be as follows:

undefined

8 0
1 year ago
Find the principal needed now to get to $25,500 after 10 years at 6% compounded continuously
Inessa05 [86]

Answer: $14239.07

Explanation:

A=P(1+I)^n

25500=P(1+0.06)^10

25500=P(1.06)^10

25500÷(1.06)^10=P

P=$14239.07

6 0
3 years ago
Which of the following lines will have a negative slope? Select all that apply
mariarad [96]
I think it is 1, 4, and 5
6 0
2 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Really need help ASAP
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The third and fourth charts represent functions
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