A rectangular sheet of paper had a width of 841 millimetres.

Linda, Tony, and deon have a total of $99 in their wallets. Tony's has 4 times what deon has. Linda has $9 more than Deon. How m

Westkost [7]

Answer:its 9 just ook the tsest

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

If each of the following questions were asked to a group of second graders, which question would NOT be a statistical question?

sweet-ann [11.9K]

Answer:

d

Step-by-step explanation:

just makes sense bc they're all in 2nd grade

8 0

2 years ago

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o

blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown. Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860. Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

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Disclaimer: This question was incomplete. Please find the full content below.

Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

#SPJ4

4 0

1 year ago

PLEASE PLEASE PLEASE HELP.

ZanzabumX [31]

Answer:

D

Step-by-step explanation:

The quotient is the result of dividing x by 5

The expression is then

15 - $\frac{x}{5}$ → D

7 0

2 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago