Quantities x and y are in a proportional relationship.

Find the odds for and the odds against the event rolling a fair die and getting a 4, a 2, or a 1.

HACTEHA [7]

Answer: Both in favor and against: 3/6 OR 1/2 OR 50%.

Step-by-step explanation: Both in favor and against: 3/6 OR 1/2 OR 50%.

The probability of rolling one die and getting any digit is $\frac{1}{6}$ . That is because the desired outcome is only 1 digit out of the 6 total digits on each of the six sides.

That means the probability of rolling either a 1, a 2, or a 4 <em>if you are looking for any of those numbers specifically, </em>is $\frac{1}{6}$ . But! You are looking to roll a 1, a 2, or a 4, without any preference among those three. Therefore, you are going to add the desired outcomes.

$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ = 0.5 = 50%

Above, are added the probability of rolling a 1 PLUS the probability of rolling a 2 PLUS the probability of rolling a 4, yielding a 50% probability.

Think about it. There are six total digits on a die. You are okay with rolling three of those. Three is half of six. Therefore, the probability of rolling the number you'd like is one-half, fifty-percent.

Likewise, since the probability of all complement events happening is 1 (or 100%), The odds against rolling either 1, 2, or 4 is also fifty-percent. 100% (which is rolling <em>any </em>number on a die) MINUS 50% (which is rolling a 1, a 2, or a 4 with no preference) EQUALS TO 50% (which is rolling a 3, a 5, or a 6).

8 0

2 years ago

A projectile is fired from a cliff 190 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of

jenyasd209 [6]

Answer:

Step-by-step explanation:

I'm not sure if this problem is physics based or calculus based. So I used calculus because it just makes more sense to do so.

If the projectile is fired from 190 feet in the air, then h0 in our quadratic will be 190. Since we are being asked to find both the displacement in the x-dimension and in the y-dimension, we need a bit of physics here as well. The velocity in the y-dimension is found in

$v_{0y}=v_{0}sin\theta$ and the velocity in the x-dimension is found in

$v_{0x}=v_{0}cos\theta$

Since the sin and the cos of 45 is the same, it's made a bit simpler for us. The velocity in both the x and the y dimension is 35.35533906 feet per second.

We can use that now to write the quadratic we need to start solving these rather tedious problems.

The position function for the projectile is

$s(t)=-16t^2+35.35533906t+190$

I'm going to kind of mix things up a bit, because in order to find the distance in the horizontal dimension that the object is when it's at its max height, we need to first find out how long it takes to get to its max height. We will first take the derivative of the position function to get the velocity function of the projectile. The first derivative of the position function is

$v(t)=-32t+35.35533906$

Remember that the first derivative is the velocity function of the projectile. You should know from either physics or calculus that at its max height, the velocity of an object is 0 (because it has to stop in the air in order to turn around and come back down). Setting the velocity function equal to 0 and solving for time will give us the time that the object is at the max height.

$0=-32t+35.35533906$

I'm going to factor out the -32 to make things easier:

$0=-32(t-1.104854346)$ which gives us that at approximately 1.10485 seconds the object is at its max height.

Moving over to the horizontal distance question now. The displacement the object experiences in the horizontal dimension is found in d = rt. We know the horizontal velocity and now we know how long it takes to get to its max height, so the horizontal distance is found in

d = (35.35533906(1.10485) so

d = 39.06 feet When the object is at its max height the object is a horizontal distance of 39.06 feet from the face of the cliff. That's a.

Now to find the max height, we will use again how long it took to get to the max height and sub it in for t in the position function.

s(1.10485) = -16(1.10485)^2 + 35.35533906(1.10485) + 190 to get that the max height is

209 feet. That's b.

Now for c. We are asked when the object will hit the water. We know that when the object hits the water it is no longer in the air and has a height of 0 above the water. Sub in a 0 for s(t) in the original position function and factor to solve for t:

$0=-16t^2+$ 35.35533906t + 190 and solve for t by factoring however you find to be the easiest. Quadratic formula works great!

We find that the times are -2.51394 seconds and 4.72365 seconds. Since time will NEVER be negative, we know that the time it takes to hit the water is 4.72365 seconds.

Whew!!!

8 0

3 years ago

How much is one troy ounce of silver worth?

alexdok [17]

One troy ounce of silver is worth, $17.67.
HOPE THIS HELPS YOU! ^_^

3 0

2 years ago

Write the denominator of the fractiona 7/20 as the product of prime factors

gtnhenbr [62]

Answer:

Step-by-step explanation:

prime factors of 20 = 2 * 2 * 5

$\frac{7}{20}=\frac{7}{2*2*5}$

5 0

2 years ago

(This question is related to the picture question)

UNO [17]

Answer:

5. $f(x) = 10,000 (1.5)^x$

Step-by-step explanation:

==>>We would have to multiply the original amount by 1.50^x because the initial amount would be 1, and 50% increase would be .5 so 1.5 and you raise it to the number of years to show the total increase.

Let's test it

Initial: 10,000

After 1 year

10,000 + (.5*10000)

10,000 + 5000 = 15,000

--

After 2 years

15,000 + (.5*15000)

15,000 + 7500 = 22,500

--

Let's try our equation.

$f(x) = 10,000 (1.5)^x$

$x=2$

$10,000(1.5)^2$

$10,000(2.25) = 22,500$

*Edit: I just seen that the question is related to the picture question

6 0

2 years ago