Ron wants to run 6 miles this week. He ran

An airplane leaves the ground in Seattle at 11:23PM on Tuesday and lands in San Francisco 1AM. Compute the average speed of the

elena55 [62]

Answer:

223.34 m/s

Step-by-step explanation:

Given :

Travel time = 11 :23 pm to 1:00 a. M = 1 hour 37 minutes = (60 + 37) = 97 minutes = (97 * 60) = 5820 seconds

Distance between Seattle and San Francisco is

807.7 miles (culled from the internet)

807.7 miles = 1299867.1 meters

Average speed = Distance covered / Time taken

Average speed = 1299867.1 m / 5820 s

Average speed = 223.34486

Average speed = 223.34 m/s

5 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

(c) The series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago

Which of the following describes the zeroes if the graph of f(x) = -x^5+9x^4-18x^3

Akimi4 [234]

Answer:

X^3= 0, X= -6,-3

Step-by-step explanation

first find the gcf(x^3)

then factor normally

7 0

3 years ago

Read 2 more answers

Fewer young people are driving. In year A, 67.9% of people under 20 years old who were eligible had a driver's license. Twenty y

aleksley [76]

Answer:

Case a Case b

margin of error 0.0216 0.0231

Interval estimate (0.7016 , 0.6795) (0.5031 , 0.4569)

margin of error is not same in both cases.

Step-by-step explanation:

a

At 95% confidence interval the interval estimate of number of 20 year old drivers in year A can be computed as

p' ± z $\sqrt{\frac{p'(1-p')}{n} }$

= 0.68 ± 1.96 $\sqrt{\frac{0.68(1-0.68)}{1800} }$

= 0.7016 , 0.6795

the margin of error can be written as

z $\sqrt{\frac{p'(1-p')}{n} }$

= 1.96 $\sqrt{\frac{0.68(1-0.68)}{1800} }$

= 0.0216

b

At 95% confidence interval the interval estimate of number of 20 year old drivers in year B can be computed as

p' ± z $\sqrt{\frac{p'(1-p')}{n} }$

= 0.48 ± 1.96 $\sqrt{\frac{0.48(1-0.48)}{1800} }$

= 0.5031 , 0.4569

the margin of error can be written as

z $\sqrt{\frac{p'(1-p')}{n} }$

= 1.96 $\sqrt{\frac{0.48(1-0.48)}{1800} }$

= 0.0231

c

Sample size is same in case A and B but proportion is different in both cases so margin of error is different in both cases

3 0

3 years ago

Ty is right inches taller than his brother Reece if ty is 42 inches tall how tall is Reece. Write an equation using a variable r

fgiga [73]

Mi llamos Bryson y tu ¿

4 0

3 years ago