The formula for this is d=rt where d= distance, r= rate of travel, and t= time
d= 3(.083) (5 minutes divided into 60 minutes per hour)
You could also convert the miles per hour to miles per minute and use the 5. 3/60= .05 miles per minute.
d= .05(5)
d= .25 miles
d= .249 miles

8 0

4 years ago

Given that f(x) = x2 + 5x + 6 and g(x) = x + 2, find (fºg)(x) and

PolarNik [594]

Answer:

$(f\circ g)(x)=x^2+9x+20$

Step-by-step explanation:

We have the two functions:

$f(x)=x^2+5x+6\text{ and } g(x)=x+2$

And we want to find:

$(f\circ g)(x)$

This is equivalent to:

$(f\circ g)(x)=f(g(x))$

Therefore, we will have:

$f(x+2)$

By substitution:

$f(x+2)=(x+2)^2+5(x+2)+6$

Simplify. Square and distribute:

$f(x+2)=(x^2+4x+4)+(5x+10)+6$

Rearrange:

$f(x+2)=(x^2)+(4x+5x)+(4+10+6)$

Combine like terms:

$f(x+2)=(f\circ g)(x)=x^2+9x+20$

7 0

3 years ago

he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

Sonja [21]

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is $\hat{p} = 199/830 = 0.24$ and an estimate of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}$ = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148$ where $z_{0.075}$ is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is $0.24\pm (1.4395)(0.0148)$ or equivalently (0.2187, 0.2613).

7 0

3 years ago