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3 years ago

X2 - 45x + 324 = 0 . middle term splitting method?

1 answer:

8 0

Factoring x2-45x+324

The first term is, x2 its coefficient is 1 .
The middle term is, -45x its coefficient is -45 .
The last term, "the constant", is +324

Step-1 : Multiply the coefficient of the first term by the constant 1 • 324 = 324

Step-2 : Find two factors of 324 whose sum equals the coefficient of the middle term, which is -45 .

-324 + -1 = -325 -162 + -2 = -164 -108 + -3 = -111 -81 + -4 = -85 -54 + -6 = -60 -36 + -9 = -45 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -36  and  -9
                     x2 - 36x - 9x - 324

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x-36)
              Add up the last 2 terms, pulling out common factors :
                    9 • (x-36)
Step-5 : Add up the four terms of step 4 :
                    (x-9)  •  (x-36)
             Which is the desired factorization

Equation at the end of step 1 : (x - 9) • (x - 36) = 0

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

Let $\mu$ = average miles for deluxe tires

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, test statistics = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

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