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3 years ago

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The number of problems assigned as homework by a savvy math professor is a function of the length of the class on a particular d

ay, which is either 45 minutes, 90 minutes, or 135 minutes. He first multiplies the number of minutes by 4 and then divides that number by 9. Finally, he subtracts 12 from that amount. Which of the following functions best represents the number of math problems the professor assigns depending on the length of the class, t, in minutes?

1 answer:

uysha [10]3 years ago

5 0

Answer:f(t)=4/9t — 12

Step-by-step explanation:

1. 4 is being multiplied by t first then divided by 9 not by 4/9 so there shouldn’t be any parentheses.

2. 12 is being subtracted from this so it should be 4/9t—12.

3. Final answer: f(t)= 4/9t — 12

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For the real-valued functions, find the composition and specify its domain using interval notation.

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Answer:

Composition: $f(x)\,^o\,h(x)=4\,\sqrt{x+4} -1$

Domain = { $x\,/\,x\geq \,-4$ }

Step-by-step explanation:

Given

$f(x)=4\,x-1\,\,\, and \,\,\, h(x)=\sqrt{x+4}$

The requested composition becomes:

$f(x)\,^o\,h(x)=f(h(x))\\f(x)\,^o\,h(x)=f(\sqrt{x+4} )\\f(x)\,^o\,h(x)=4\,\sqrt{x+4} -1$

Then the Domain consists of all Real x-values for which the function is defined. In this case, the function is defined for all x-values for which the square root is defined, that is all x values for which:

$x+4\geq \,0$ (so the square root is defined in the Real number system)

This means $x\geq \,-4$

The Domain is then all the Real x-values such that $x\geq \,-4$

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3 years ago

Discuss on piazza but do not submit. Divide [-2,2] into (a) 10 subintervals (b) 1000 subintervals of the same length. Then use t

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Answer:

attached below

Step-by-step explanation:

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3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

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Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ; </u>

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u> 95% confidence interval for</u> $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions</em>

<u>So, 99% confidence interval for the population proportion, p is ; </u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u> 99% confidence interval for</u> p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

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