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Alenkasestr [34]
3 years ago
5

Which of the following is the correct factored form of the given equation?

Mathematics
2 answers:
katrin [286]3 years ago
6 0

Hint: "4x 2 - 25 = 0" is nearly meaningless. You want 4x^2 - 25 = 0.

Note that both 4x^2 and 25 are perfect squares, so this is the difference of two squares, and is equal to (2x-5)(2x+5)= 0.

dmitriy555 [2]3 years ago
4 0

There is a small subset of polynomials whos factorizations mathematicians like to label "special." From an outside perspective, nothing seems to be different about these polynomials. You can use the same algorithm you use to find any other factorization, but you could also form a general statement about polynomials like this.

This polynomial is in the form a^2-b^2. We can factor this normally.

-b and b add to 0 and multiply to -b^2, so:

a^2+ab-ab-b^2=\\ a(a+b)-b(a-b)=\\ (a-b)(a+b)

So, we've factored this quadratic like we would any other, but we also notice something interesting about all quadratics in this form. In this example,

a^2=4x^2\\ b^2=25

That means

a=2x\\ b=5

So we can say that the factorization of the quadratic is:

(2x-5)(2x+5)

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• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

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Answer:

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