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2 years ago

The common factors of 32 and 72 are

Mathematics

2 answers:

ivanzaharov [21]2 years ago

8 0

Answer:

The common factor of 32 and 72 are-

1,2,4, and 8

marusya05 [52]2 years ago

8 0

The answer is letter b

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2 years ago

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Find the area of the shaded region. round your answer to the nearest hundredth

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Answer:

16 meters

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3 years ago

What’s the answer ?

xxMikexx [17]

B 5x+20 hope this helps!!

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3 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.

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:

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

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

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

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.

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

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

For 300 trading days, the daily closing price of a stock (in $) is well modeled by a Normal model with mean $196.99 and stand

KATRIN_1 [288]

Answer:

a) 2.5% b) 84% c) 95% d) D. The more unusual day is if the stock closed below $185 because it has the largest absolute z-score.

Step-by-step explanation:

For a) b) and c) we will use the empirical rule, so, we can observe the image shown below

a) 211.23 is exactly two standard deviation above the mean, so, the probability that on a randomly selected day in this period the stock price closed above 211.23 is 2.35% + 0.15% = 2.5%

b) 204.11 represents exactly one standard deviation above the mean, so, the probability of being below 204.11 is 50% + 34% = 84%

c) The probability of getting a value between 182.75 and 211.23 is 95%, this because 182.75 is exactly two standard deviations below the mean and 211.23 is exactly two standard deviations above the mean.

d) The z-score related to 208 is $z_{1}$ = (208-196.99)/7.12 = 1.5 and the z-score related to 185 is $z_{2}$ = (185-196.99)/7.12 = -1.7, therefore, the more unusual day is if the stock closed below $185 because it has the largest absolute z-score.

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3 years ago