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nignag [31]
2 years ago
11

Compare and Contrast how you would factor x^2−9 and x^2+9. For full credit, make sure you explain why they are factored differen

tly.
Mathematics
1 answer:
marishachu [46]2 years ago
7 0

The factored form of the expression x^2 - 9 is (x+3)(x-3)

The expression x^2 + 9 has no real roots hence cannot be factored

Quadratic equations are equations with a degree of 2.

Given the quadratic function x^2-9 \ and \ x^2+9

For the function x^2 - 9

=x^2-9\\=x^2-3^2

Using the difference of two squares:

a^2-b^2=(a-b)(a+b)\\x^2-3^2=(x-3)(x+3)

For the expression x^2 + 9

The expression x^2 + 9 has no real roots hence cannot be factored

Learn more on factorization here: brainly.com/question/20293447

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Rewrite the expression as a multiple of a sum of two numbers with no common factor. <br> 30 + 12
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A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco
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Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

            n = sample of cars = 600

So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

                               =  -0.866

<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

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