Questions: Factor the trinomial. 36 p^3 + 168 p^2 w + 196 p w^2 =

Factor the trinomial.
36 p^3 + 168 p^2 w + 196 p w^2 =

Transcript text: Factor the trinomial. \[ 36 p^{3}+168 p^{2} w+196 p w^{2}= \]

Solution

Solution Steps

To factor the given trinomial \(36p^3 + 168p^2w + 196pw^2\), we can follow these high-level steps:

Identify the greatest common factor (GCF) of the coefficients.
Factor out the GCF from the trinomial.
Check if the remaining trinomial can be factored further, possibly as a perfect square trinomial.

Step 1: Identify the Greatest Common Factor (GCF)

First, we identify the greatest common factor (GCF) of the coefficients in the trinomial \(36p^3 + 168p^2w + 196pw^2\). The GCF of 36, 168, and 196 is 4.

Step 2: Factor out the GCF

Next, we factor out the GCF from the trinomial: \[ 36p^3 + 168p^2w + 196pw^2 = 4(9p^3 + 42p^2w + 49pw^2) \]

Step 3: Factor the Remaining Trinomial

We observe that the remaining trinomial \(9p^3 + 42p^2w + 49pw^2\) can be factored further. It is a perfect square trinomial: \[ 9p^3 + 42p^2w + 49pw^2 = (3p + 7w)^2 \]

Final Answer

Combining the factored GCF and the factored trinomial, we get: \[ 36p^3 + 168p^2w + 196pw^2 = 4p(3p + 7w)^2 \] \[ \boxed{4p(3p + 7w)^2} \]

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The TTC has set a bus mechanical reliability goal of 3,900 bus miles. Bus mechanical reliability is measured specifically as the number of bus miles between mechanical road calls. Suppose a sample of 100 buses resulted in a sample mean of 3,950 bus miles and a sample standard deviation of 175 bus miles. Complete parts (a) and (b) below. a. Is there evidence that the population mean bus miles is more than 3,900 bus miles? (Use a 0.05 level of significance.) State the null and alternative hypotheses. H0: μ = T H1: μ ≠ T (Type integers) Find the test statistic for this hypothesis test. The test statistic t= (Round to two decimal places as needed.) The critical value for the test statistic is(are) . (Round to two decimal places as needed) Is there sufficient evidence to reject the null hypothesis using α=0.05 ? A. Do not reject the null hypothesis. There is insufficient evidence at the 0.05 level of significance that the population mean bus miles is less than 3,900 bus miles. B. Reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance that the population mean bus miles is greater than 3,900 bus miles. C. Reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance that the population mean bus miles is less than 3,900 bus miles. D. Do not reject the null hypothesis. There is insufficient evidence at the 0.05 level of significance that the population mean bus miles is greater than 3,900 bus miles. b. The p-value is . (Round to 3 decimal places as needed.) What does this p-value mean given the results of part (a)? A. The p-value is the probability of getting a sample mean of 3,950 bus miles or greater if the actual mean is 3,900 bus miles. B. The p-value is the probability that the actual mean is 3,950 bus miles or less. C. The p-value is the probability that the actual mean is 3,900 bus miles or greater given the sample mean is 3,950 bus miles.

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