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jenyasd209 [6]
3 years ago
7

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

, 2, 3, 4 f'(x): 74, 39, 10, -6, -14, -12, 0, 22, 55 a) estimate where f is increasing, decreasing, and has local extrema.
Mathematics
1 answer:
kondaur [170]3 years ago
4 0
f will be increasing on the intervals where f'(x)>0 and decreasing wherever f'(x). Local extrema occur when f'(x)=0 and the sign of f'(x) changes to either side of that point.

f'(x) is positive when x is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that f(x) is increasing on the intervals [-4, -2] and (2, 4].

f'(x) is negative when x is between some number between -2 and -1, up to some number less than 2. So f(x) is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of f'(x) changes for some x between -2 and -1, and again to either side of x=2.
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The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 13001300
Ket [755]

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No, there is not enough evidence at the 0.02 level to support the strategist's claim.

Step-by-step explanation:

We are given that a political study took a sample of 1300 voters in the town and found that 57% of the residents favored construction.

And, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 54%, i.e;

Null Hypothesis, H_0 : p = 0.54 {means that the percentage of residents who favor construction is 54%}

Alternate Hypothesis, H_1 : p > 0.54 {means that the percentage of residents who favor construction is more than 54%}

The test statistics we will use here is;

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

where, p = actual percentage of residents who favor construction = 0.54

           \hat p = percentage of residents who favor construction in a sample of

                  1300 voters = 0.57

           n = sample of voters = 1300

So, Test statistics = \frac{0.57 -0.54}{\sqrt{\frac{0.57(1- 0.57)}{1300} } }

                              = 2.185

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the percentage of residents who favor construction is 54% and the strategist's claim is not supported.

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