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andreyandreev [35.5K]

2 years ago

14

There are 26 baseball teams in the league.Each team has 18 players.Whats a number sentence that will provide a reasonable estima

te for the number of players in the league

1 answer:

klio [65]2 years ago

6 0

The anwser is along the lines of "there are 26 teams in the league, and if each team has 18 players then 26 times 18 is equal to 468 players in the league.
hope this helps and dont forget to hit that heart :)

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Find the range of the given data set.<br><br> -4, -3, -1, -1, 0, 1

Dovator [93]

Answer:

5

Step-by-step explanation:

To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

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3 years ago

The question is show a dilation of a rectangle with a dilation of 8/3

miv72 [106K]

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<span><span>1) multiply both coordinates by scale factor<span>(<span><span>2⋅<span>12</span>,4⋅<span>12</span></span><span>2⋅<span>12</span>,4⋅<span>12</span></span></span>)</span></span><span>2)

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3 years ago

Each vertical cross-section of the triangular

Blizzard [7]

Answer:

The vertical height is $h=3\ units$

Step-by-step explanation:

we know that

In the right triangle ABC

see the attached figure

To find the height h apply the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

we have

$c=AB=3\sqrt{2}\ units$

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substitute

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3 0

2 years ago

Read 2 more answers

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

A box contains eight cards labeled P, Q, R, S, T, U, V, and W. One card will be randomly chosen.

Varvara68 [4.7K]

Answer:

7/8

Step-by-step explanation:

Probability= (Cards having letter Q to W)/(All cards) = 7/8

6 0

3 years ago