Does this graph represent a function why or why not

In a study of academic procrastination, the authors of a paper reported that for a sample of 471 undergraduate students at a mid

AnnZ [28]

Answer:

95% confidence interval for the mean time spent studying for the final exam for students taking introductory psychology at this university is [6.85 hours, 7.43 hours].

Step-by-step explanation:

We are given that a sample of 471 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.14 hours and the standard deviation of study times was 3.20 hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean time spent studying for the exam = 7.14 hours

$\sigma$ = population standard deviation = 3.20 minutes

n = sample of undergraduate students = 471

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $7.14-1.96 \times {\frac{3.20}{\sqrt{471} } }$ , $7.14+1.96 \times {\frac{3.20}{\sqrt{471} } }$ ]

= [6.85 hours , 7.43 hours]

Therefore, 95% confidence interval for the mean time spent studying for the final exam for students taking introductory psychology at this university is [6.85 hours, 7.43 hours].

7 0

3 years ago

Yeooo guys please help me :) !

Afina-wow [57]

The answer is $138.98 Dollars

6 0

3 years ago

Evaluate 8+6÷2+(2+10) help and work with answer please

Svetradugi [14.3K]

Answer:

23

Step-by-step explanation:

P E M D A S

8+6÷2+(2+10)

P)

8+6÷2+(2+10)

8+6÷2+(12)

E)

M)

D)

8+6÷2+(12)

8+3+12

A)

8+3+12

11+12

S)

The final answer is 23.

3 0

3 years ago

Read 2 more answers

Can y’all help me answer these questions

cupoosta [38]

2. $f(x)=\dfrac18\left(\dfrac14\right)^{x-2}$

Domain: $(-\infty,\infty)$ , because any value of $x$ is allowed and gives a number $f(x)$ .

Range: $(0,\infty)$ , because $a^x>0$ for any positive real $a\neq0$ .

y-intercept: This is a point of the form $(0, f(0))$ . So plug in $x=0$ ; we get $f(0)=\frac18\left(\frac14\right)^{-2}=\frac{4^2}8=2$ . So the intercept is (0, 2), or just 2. (Interestingly, you didn't get marked wrong for that...)

Asymptote: This can be deduced from the range; the asymptote is the line $y=0$ .

Increasing interval: Going from left to right, there is no interval on which $f(x)$ is increasing, since 1/4 is between 0 and 1.

Decreasing interval: Same as the domain; $f(x)$ is decreasing over the entire real line.

End behavior: The range tells you $f(x)>0$ , and you know $f(x)$ is decreasing over its entire domain. This means that $f(x)\to+\infty$ as $x\to-\infty$ , and $f(x)\to0$ and $x\to+\infty$ .

3. $f(x)=\left(\dfrac32\right)^{-x}-7$

Domain: Same as (2), $(-\infty, \infty)$ .

Range: We can rewrite $f(x)=\left(\frac23\right)^x-7$ . $\left(\frac23\right)^x>0$ for all $x$ , so $\left(\frac23\right)^x-7>-7$ for all $x$ . Then the range is $(-7,\infty)$ .