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2 years ago

Jason tossed a fair coin 3 times. what is the probability of getting a head and two tails in any order?

Mathematics

1 answer:

mixer [17]2 years ago

4 0

Step-by-step explanation:

fair coin is two sided Head H and Tail T

thrown 3 times

number of sample space n(s)=2³=>8

sample space (s)={HHH,HHT,HTH,HTT,THH,TTH,THT,TTT}

no of a head and two tails= 3/8

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HELP ASAP. a kite has diagonals of 18 inches and 21 inches as seen below. what is the perimeter of the kite

timofeeve [1]

Answer:

78 inches

Step-by-step explanation:

perimeter of a kite formula: 2(a+b)

a= 18 inches

b= 21 inches

2(18+21)

2(39)

78 inches

8 0

3 years ago

What is the equation of the line parallel to 3x+2y= -4 that goes through the point (4,-1)

sertanlavr [38]

Answer:

y = (-3/2)x + 7

Step-by-step explanation:

3x + 2y = -4 (rearrange to slope intercept form y = mx + b)

2y = -3x - 4

y = (-3/2) x - 2

comparing this to the general form of a linear equation : y = mx + b

we see that slope of this line (and every line that is parallel to this line),

m = -3/2

if we sub this back in to the general form, we get:

y = (-3/2)x + b

We are still missing the value of b. To find this, we are given that the point (4,1) lies on the line. We simply substitute this back into the equation and solve for b.

1 = (-3/2)4 + b

1 = -6 + b

b = 7

substituting this back into the equation:

y = (-3/2)x + 7

6 0

3 years ago

Read 2 more answers

A bottle can hold 2.75 liters of water. About how many gallons can the bottle hold? Round to the nearest hundredth.

Serjik [45]

1 gallon is equal to 3.78541 liters

6 0

3 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.

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

Decide whether the following rates are equivalent. Show work; and answer of YES or NO.

zubka84 [21]

Eh 12/20 is .6 pages a minute. 20/32 is .625 pages a minute

8 0

3 years ago