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egoroff_w [7]
3 years ago
12

Identify each x-value at which the slope of the tangent line to the function f(x) = 0.2x^2 + 5x − 12 belongs to the interval (-1

, 1).

Mathematics
2 answers:
zhuklara [117]3 years ago
5 0

Answer:

Step-by-step explanation:

Hello, the slope of the tangent is the value of the derivative.

f'(x) = 2*0.2x + 5 = 0.4x + 5

So we are looking for

-1\leq f'(x) \leq 1 \\ \\ -1\leq 0.4x+5 \leq 1 \\ \\ -1-5=-6\leq 0.4x \leq 1-5=-4 \\ \\ \dfrac{-6}{0.4}\leq 0.4x \leq \dfrac{-4}{0.4} \\\\ \boxed{-15 \leq x\leq -10}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

DIA [1.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

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(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms i.e.;

Sample mean, xbar = 290      Sample standard deviation, s = 30  and  Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

                         \sigma^{2}  =  s^{2} = 30^{2}

                          \sigma^{2}  = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of  \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

P(10.12 < \chi^{2}__1_9 < 30.14) = 0.90 {At 10% significance level chi square has critical

                                           values of 10.12 and 30.14 at 19 degree of freedom}        

P(10.12 < \frac{(n-1)s^{2} }{\sigma^{2} } < 30.14) = 0.90

P(\frac{10.12}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{30.14}{(n-1)s^{2} } ) = 0.90

P(\frac{(n-1)s^{2} }{30.14} < \sigma^{2} < \frac{(n-1)s^{2} }{10.12} ) = 0.90

90% confidence interval for \sigma^{2} = [\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]

                                                   = [\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]

                                                   = [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

(c) 90% confidence interval estimate of the population standard deviation is given by ;

       P(\sqrt{\frac{(n-1)s^{2} }{30.14}} < \sigma < \sqrt{\frac{(n-1)s^{2} }{10.12}} ) = 0.90

90% confidence interval for \sigma = [\sqrt{\frac{19s^{2} }{30.14}}   , \sqrt{\frac{19s^{2} }{10.12}}  ]

                                                 = [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .

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