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choli [55]
3 years ago
5

A communications company has developed three new designs for a cell phone. To evaluate consumer response, a sample of n = 120 co

llege students is selected and each student is given all three phones to use for one week. At the end of the week, the students must identify which of the three designs they prefer. The distribution of preference is as follows: Design 1=54, Design 2=38, Design 3=28. Do the results indicate any significant preferences among the three designs?
Mathematics
1 answer:
defon3 years ago
4 0

Answer:

The answer is "Its results are not consistent with its three designs Yes. Its results show significant differences between both the three designs."

Step-by-step explanation:

Following are the distribution of preference:  

Design 1 =54 \\\\ Design 2=38 \\\\\ Design 3=28 \\\\

Design \ \ \ \ \ \ \ \ \ \ O_i \\ 1 \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ 54 \\2  \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ 38\\3 \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  28 \\Total \ \ \ \ \ \ \ \ \ \ 120

An expected frequency(E_i) has 40 Null hypotheses to take n=120.to each design:  

The effects of the three designs are standardized  

Inaccurate.  

Hypothesis Alternative:  

The effects of the three prototypes are not uniform

Degrees Of freedom =(n-1)=(3-1)=2

H_o, x^2  \ test\  statistic \  \sum_i {\frac{(O_i -E_i)^2}{E_i}} \\

= \frac{(54 -40)^2}{40}+\frac{(38-40)^2}{40} + \frac{(28-40)^2}{40}\\\\=49+0.1+3.6\\\\=8.6

Chi-squared distribution critical value at \alpha = 0.05 (x^2 _{0.052})=5.99  

Since x^2, a value > x^2 -table of \alpha =0.05 =5.99is determined.  

SoH_o is rejected.

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stepan [7]

Answer:

Factors: (x²y+6)(y²-11)

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Step-by-step explanation:

Here's the original equation.

x²y³ - 11x²y + 6y² - 66

Let's separate it into two groups to make it easier.

(x²y³ - 11x²y) + (6y² - 66)

Let's take a look at (x²y³ - 11x²y) first. Do we notice anything that can be removed from the parentheses?

Yes, x²y can be taken from both polynomials to simplify the contents of the parentheses.

Both polynomials have x² and both have at least one y. Combine that to get x²y. That means that within the parentheses y² and - 11 will be left.

So  (x²y³ - 11x²y) becomes x²y (y²-11)

This works because if we multiply x²y * (y² - 11) we get (x²y³ - 11x²)

Let's apply this same idea to (6y² - 66).

6 is the only thing in common between the two polynomials.

If we follow the same steps we get 6 (y²-11).

Notice how when we simplified both groups they each had something on the outside and (y² - 11) in the parentheses?

So lets combine the things that were on the outside of both (y² - 11).

What was on the outside of each? x²y was on the outside for the first one, and 6 for the outside of the second one.

This means that if we combine everything and simplify, the factors we have are (x²y+6)(y²-11).

If you want to confirm, work your way backwards using the distributive property. You'll find that when you multiply (x²y+6) * (y² -11) you'll get = x²y³ - 11x²y + 6y² - 66. That was the original equation, so we know that these are the factors.

Hope that helps!

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Answer:

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Step-by-step explanation:

d = sqrt [(x-x)^2 + (y-y)^2]

d = sqrt [(-7 - - 2)^2 + (7-3)^2]

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